Physics 12th Book 1,2

Physics FIGURE 11.1 Experimental attracted to it. The emission of electrons causes flow of arrangement for study of electric current in the circuit. The potential difference photoelectric effect. between the emitter and collector plates is measured by a voltmeter (V) whereas the resulting photo current flowing in the circuit is measured by a microammeter (µA). The photoelectric current can be increased or decreased by varying the potential of collector plate A with respect to the emitter plate C. The intensity and frequency of the incident light can be varied, as can the potential difference V between the emitter C and the collector A. We can use the experimental arrangement of Fig. 11.1 to study the variation of photocurrent with (a) intensity of radiation, (b) frequency of incident radiation, (c) the potential difference between the plates A and C, and (d) the nature of the material of plate C. Light of different frequencies can be used by putting appropriate coloured filter or coloured glass in the path of light falling on the emitter C. The intensity of light is varied by changing the distance of the light source from the emitter. FIGURE 11.2 Variation of 11.4.1 Effect of intensity of light on photocurrent Photoelectric current with The collector A is maintained at a positive potential with intensity of light. respect to emitter C so that electrons ejected from C are attracted towards collector A. Keeping the frequency of the incident radiation and the potential fixed, the intensity of light is varied and the resulting photoelectric current is measured each time. It is found that the photocurrent increases linearly with intensity of incident light as shown graphically in Fig. 11.2. The photocurrent is directly proportional to the number of photoelectrons emitted per second. This implies that the number of photoelectrons emitted per second is directly proportional to the intensity of incident radiation. 11.4.2 Effect of potential on photoelectric current We first keep the plate A at some positive potential with respect to the plate C and illuminate the plate C with light of fixed frequency ν and fixed intensity I1. We next vary the positive potential of plate A gradually and measure the resulting photocurrent each time. It is found that the photoelectric current increases with increase in positive (accelerating) potential. At some stage, for a certain positive potential of plate A, all the emitted electrons are collected by the plate A and the photoelectric current becomes maximum or saturates. If we increase the accelerating potential of plate A further, the photocurrent does not increase. This maximum value of the photoelectric current is called saturation current. Saturation current corresponds to the case when all the photoelectrons emitted by the emitter plate C reach the collector plate A. We now apply a negative (retarding) potential to the plate A with respect 390 to the plate C and make it increasingly negative gradually. When the 2020-21

Dual Nature of Radiation and Matter polarity is reversed, the electrons are repelled and only the sufficiently energetic electrons are able to reach the collector A. The photocurrent is found to decrease rapidly until it drops to zero at a certain sharply defined, critical value of the negative potential V0 on the plate A. For a particular frequency of incident radiation, the minimum negative (retarding) potential V0 given to the plate A for which the photocurrent stops or becomes zero is called the cut-off or stopping potential. The interpretation of the observation in terms of photoelectrons is straightforward. All the photoelectrons emitted from the metal do not have the same energy. FIGURE 11.3 Variation of photocurrent with Photoelectric current is zero when the collector plate potential for different stopping potential is sufficient to repel even intensity of incident radiation. the most energetic photoelectrons, with the maximum kinetic energy (Kmax), so that Kmax = e V0 (11.1) We can now repeat this experiment with incident radiation of the same frequency but of higher intensity I2 and I3 (I3 > I2 > I1). We note that the saturation currents are now found to be at higher values. This shows that more electrons are being emitted per second, proportional to the intensity of incident radiation. But the stopping potential remains the same as that for the incident radiation of intensity I1, as shown graphically in Fig. 11.3. Thus, for a given frequency of the incident radiation, the stopping potential is independent of its intensity. In other words, the maximum kinetic energy of photoelectrons depends on the light source and the emitter plate material, but is independent of intensity of incident radiation. 11.4.3 Effect of frequency of incident radiation on stopping potential We now study the relation between the FIGURE 11.4 Variation of photoelectric current frequency ν of the incident radiation and the with collector plate potential for different stopping potential V0. We suitably adjust the frequencies of incident radiation. same intensity of light radiation at various frequencies and study the variation of 391 photocurrent with collector plate potential. The resulting variation is shown in Fig. 11.4. We obtain different values of stopping potential but the same value of the saturation current for incident radiation of different frequencies. The energy of the emitted electrons depends on the frequency of the incident radiations. The stopping potential is more negative for higher frequencies of incident radiation. Note from 2020-21

Physics Fig. 11.4 that the stopping potentials are in the order V03 > V02 > V01 if the frequencies are in the order ν3 > ν2 > ν1 . This implies that greater the frequency of incident light, greater is the maximum kinetic energy of the photoelectrons. Consequently, we need greater retarding potential to stop them completely. If we plot a graph between the frequency of incident radiation and the corresponding stopping potential for different metals we get a straight line, as shown FIGURE 11.5 Variation of stopping potential V0 in Fig. 11.5. with frequency ν of incident radiation for a The graph shows that given photosensitive material. (i) the stopping potential V0 varies linearly with the frequency of incident radiation for a given photosensitive material. (ii) there exists a certain minimum cut-off frequency ν0 for which the stopping potential is zero. These observations have two implications: (i) The maximum kinetic energy of the photoelectrons varies linearly with the frequency of incident radiation, but is independent of its intensity. (ii) For a frequency ν of incident radiation, lower than the cut-off frequency ν0, no photoelectric emission is possible even if the intensity is large. This minimum, cut-off frequency ν0, is called the threshold frequency. It is different for different metals. Different photosensitive materials respond differently to light. Selenium is more sensitive than zinc or copper. The same photosensitive substance gives different response to light of different wavelengths. For example, ultraviolet light gives rise to photoelectric effect in copper while green or red light does not. Note that in all the above experiments, it is found that, if frequency of the incident radiation exceeds the threshold frequency, the photoelectric emission starts instantaneously without any apparent time lag, even if the incident radiation is very dim. It is now known that emission starts in a time of the order of 10– 9 s or less. We now summarise the experimental features and observations described in this section. (i) For a given photosensitive material and frequency of incident radiation (above the threshold frequency), the photoelectric current is directly proportional to the intensity of incident light (Fig. 11.2). (ii) For a given photosensitive material and frequency of incident radiation, saturation current is found to be proportional to the intensity of incident radiation whereas the stopping potential is independent of its intensity (Fig. 11.3). (iii) For a given photosensitive material, there exists a certain minimum cut-off frequency of the incident radiation, called the threshold frequency, below which no emission of photoelectrons takes place, 392 no matter how intense the incident light is. Above the threshold frequency, the stopping potential or equivalently the maximum kinetic 2020-21

Dual Nature of Radiation and Matter energy of the emitted photoelectrons increases linearly with the 393 frequency of the incident radiation, but is independent of its intensity (Fig. 11.5). (iv) The photoelectric emission is an instantaneous process without any apparent time lag (∼10– 9s or less), even when the incident radiation is made exceedingly dim. 11.5 PHOTOELECTRIC EFFECT AND WAVE THEORY OF LIGHT The wave nature of light was well established by the end of the nineteenth century. The phenomena of interference, diffraction and polarisation were explained in a natural and satisfactory way by the wave picture of light. According to this picture, light is an electromagnetic wave consisting of electric and magnetic fields with continuous distribution of energy over the region of space over which the wave is extended. Let us now see if this wave picture of light can explain the observations on photoelectric emission given in the previous section. According to the wave picture of light, the free electrons at the surface of the metal (over which the beam of radiation falls) absorb the radiant energy continuously. The greater the intensity of radiation, the greater are the amplitude of electric and magnetic fields. Consequently, the greater the intensity, the greater should be the energy absorbed by each electron. In this picture, the maximum kinetic energy of the photoelectrons on the surface is then expected to increase with increase in intensity. Also, no matter what the frequency of radiation is, a sufficiently intense beam of radiation (over sufficient time) should be able to impart enough energy to the electrons, so that they exceed the minimum energy needed to escape from the metal surface . A threshold frequency, therefore, should not exist. These expectations of the wave theory directly contradict observations (i), (ii) and (iii) given at the end of sub-section 11.4.3. Further, we should note that in the wave picture, the absorption of energy by electron takes place continuously over the entire wavefront of the radiation. Since a large number of electrons absorb energy, the energy absorbed per electron per unit time turns out to be small. Explicit calculations estimate that it can take hours or more for a single electron to pick up sufficient energy to overcome the work function and come out of the metal. This conclusion is again in striking contrast to observation (iv) that the photoelectric emission is instantaneous. In short, the wave picture is unable to explain the most basic features of photoelectric emission. 11.6 EINSTEIN’S PHOTOELECTRIC EQUATION: ENERGY QUANTUM OF RADIATION In 1905, Albert Einstein (1879 -1955) proposed a radically new picture of electromagnetic radiation to explain photoelectric effect. In this picture, photoelectric emission does not take place by continuous absorption of energy from radiation. Radiation energy is built up of discrete units – the so called quanta of energy of radiation. Each quantum of radiant energy 2020-21

Physics has energy hν, where h is Planck’s constant and ν the frequency of light. In photoelectric effect, an electron absorbs a quantum of energy (hν ) of radiation. If this quantum of energy absorbed exceeds the minimum energy needed for the electron to escape from the metal surface (work function φ0), the electron is emitted with maximum kinetic energy Kmax = hν – φ0 (11.2) More tightly bound electrons will emerge with kinetic energies less than the maximum value. Note that the intensity of light of a given frequency is determined by the number of photons incident per second. Increasing Albert Einstein (1879 – the intensity will increase the number of emitted electrons 1955) Einstein, one of the per second. However, the maximum kinetic energy of the greatest physicists of all emitted photoelectrons is determined by the energy of each time, was born in Ulm, Germany. In 1905, he photon. published three path- Equation (11.2) is known as Einstein’s photoelectric breaking papers. In the equation. We now see how this equation accounts in a first paper, he introduced simple and elegant manner all the observations on the notion of light quanta photoelectric effect given at the end of sub-section 11.4.3. (now called photons) and used it to explain the • According to Eq. (11.2), Kmax depends linearly on ν, features of photoelectric and is independent of intensity of radiation, in effect. In the second paper, agreement with observation. This has happened he developed a theory of because in Einstein’s picture, photoelectric effect arises Brownian motion, from the absorption of a single quantum of radiation confirmed experimentally a few years later and provided by a single electron. The intensity of radiation (that is a convincing evidence of proportional to the number of energy quanta per unit the atomic picture of matter. area per unit time) is irrelevant to this basic process. The third paper gave birth • Since Kmax must be non-negative, Eq. (11.2 ) implies to the special theory of that photoelectric emission is possible only if ALBERT EINSTEIN (1879 – 1955) relativity. In 1916, he published the general h ν > φ0 theory of relativity. Some of or ν > ν0 , where Einstein’s most significant ν0 = φ0 (11.3) later contributions are: the h notion of stimulated emission introduced in an Equation (11.3) shows that the greater the work alternative derivation of function φ0, the higher the minimum or threshold Planck’s blackbody frequency ν0 needed to emit photoelectrons. Thus, radiation law, static model there exists a threshold frequency ν0 (= φ0/h) for the of the universe which metal surface, below which no photoelectric emission started modern cosmology, quantum statistics of a gas is possible, no matter how intense the incident of massive bosons, and a radiation may be or how long it falls on the surface. critical analysis of the foundations of quantum • In this picture, intensity of radiation as noted above, mechanics. In 1921, he was is proportional to the number of energy quanta per awarded the Nobel Prize in unit area per unit time. The greater the number of physics for his contribution energy quanta available, the greater is the number of to theoretical physics and electrons absorbing the energy quanta and greater, the photoelectric effect. therefore, is the number of electrons coming out of 394 the metal (for ν > ν0). This explains why, for ν > ν0 , photoelectric current is proportional to intensity. 2020-21

Dual Nature of Radiation and Matter • In Einstein’s picture, the basic elementary process involved in photoelectric effect is the absorption of a light quantum by an electron. This process is instantaneous. Thus, whatever may be the intensity i.e., the number of quanta of radiation per unit area per unit time, photoelectric emission is instantaneous. Low intensity does not mean delay in emission, since the basic elementary process is the same. Intensity only determines how many electrons are able to participate in the elementary process (absorption of a light quantum by a single electron) and, therefore, the photoelectric current. Using Eq. (11.1), the photoelectric equation, Eq. (11.2), can be written as e V0 = h ν – φ 0; for ν ≥ ν0 or V0 =  h  ν − φ0 (11.4)  e  e This is an important result. It predicts that the V0 versus ν curve is a straight line with slope = (h/e), independent of the nature of the material. During 1906-1916, Millikan performed a series of experiments on photoelectric effect, aimed at disproving Einstein’s photoelectric equation. He measured the slope of the straight line obtained for sodium, similar to that shown in Fig. 11.5. Using the known value of e, he determined the value of Planck’s constant h. This value was close to the value of Planck’s contant (= 6.626 × 10–34J s) determined in an entirely different context. In this way, in 1916, Millikan proved the validity of Einstein’s photoelectric equation, instead of disproving it. The successful explanation of photoelectric effect using the hypothesis of light quanta and the experimental determination of values of h and φ0, in agreement with values obtained from other experiments, led to the acceptance of Einstein’s picture of photoelectric effect. Millikan verified photoelectric equation with great precision, for a number of alkali metals over a wide range of radiation frequencies. 11.7 PARTICLE NATURE OF LIGHT: THE PHOTON 395 Photoelectric effect thus gave evidence to the strange fact that light in interaction with matter behaved as if it was made of quanta or packets of energy, each of energy h ν. Is the light quantum of energy to be associated with a particle? Einstein arrived at the important result, that the light quantum can also be associated with momentum (h ν/c ). A definite value of energy as well as momentum is a strong sign that the light quantum can be associated with a particle. This particle was later named photon. The particle-like behaviour of light was further confirmed, in 1924, by the experiment of A.H. Compton (1892-1962) on scattering of X-rays from electrons. In 1921, Einstein was awarded the Nobel Prize in Physics for his contribution to theoretical physics and the photoelectric effect. In 1923, Millikan was awarded the Nobel Prize in physics for his work on the elementary charge of electricity and on the photoelectric effect. We can summarise the photon picture of electromagnetic radiation as follows: 2020-21

Physics (i) In interaction of radiation with matter, radiation behaves as if it is made up of particles called photons. (ii) Each photon has energy E (=hν) and momentum p (= h ν/c), and speed c, the speed of light. (iii) All photons of light of a particular frequency ν, or wavelength λ, have the same energy E (=hν = hc/λ) and momentum p (= hν/c = h/λ), whatever the intensity of radiation may be. By increasing the intensity of light of given wavelength, there is only an increase in the number of photons per second crossing a given area, with each photon having the same energy. Thus, photon energy is independent of intensity of radiation. (iv) Photons are electrically neutral and are not deflected by electric and magnetic fields. (v) In a photon-particle collision (such as photon-electron collision), the total energy and total momentum are conserved. However, the number of photons may not be conserved in a collision. The photon may be absorbed or a new photon may be created. Example 11.1 Monochromatic light of frequency 6.0 ×1014 Hz is produced by a laser. The power emitted is 2.0 ×10–3 W. (a) What is the energy of a photon in the light beam? (b) How many photons per second, on an average, are emitted by the source? EXAMPLE 11.1 Solution (a) Each photon has an energy E = h ν = ( 6.63 ×10–34 J s) (6.0 ×1014 Hz) = 3.98 × 10–19 J (b) If N is the number of photons emitted by the source per second, the power P transmitted in the beam equals N times the energy per photon E, so that P = N E. Then N= P = 2.0 ×10−3 W E 3.98 ×10−19 J = 5.0 ×1015 photons per second. Example 11.2 The work function of caesium is 2.14 eV. Find (a) the threshold frequency for caesium, and (b) the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of 0.60 V. Solution (a) For the cut-off or threshold frequency, the energy h ν0 of the incident radiation must be equal to work function φ0, so that ν0 = φ0 = 2.14 eV h 6.63 ×10−34 J s EXAMPLE 11.2 = 2.14 × 1.6 × 10−19 J = 5.16 × 1014 Hz 6.63 × 10−34 J s Thus, for frequencies less than this threshold frequency, no photoelectrons are ejected. (b) Photocurrent reduces to zero, when maximum kinetic energy of 396 the emitted photoelectrons equals the potential energy e V0 by the retarding potential V0. Einstein’s Photoelectric equation is 2020-21

Dual Nature of Radiation and Matter eV0 = hν – φ 0 = hc – φ0 λ or, λ = hc/(eV0 + φ0 ) = (6.63 × 10−34 J s) × (3 × 108 m/s) (0.60 eV + 2.14 eV) EXAMPLE 11.2 = 19.89 ×10−26 J m (2.74 eV) λ = 19.89 × 10−26 J m = 454 nm 2.74 × 1.6 × 10−19 J Example 11.3 The wavelength of light in the visible region is about 390 nm for violet colour, about 550 nm (average wavelength) for yellow- green colour and about 760 nm for red colour. (a) What are the energies of photons in (eV) at the (i) violet end, (ii) average wavelength, yellow-green colour, and (iii) red end of the visible spectrum? (Take h = 6.63×10–34 J s and 1 eV = 1.6×10 –19J.) (b) From which of the photosensitive materials with work functions listed in Table 11.1 and using the results of (i), (ii) and (iii) of (a), can you build a photoelectric device that operates with visible light? Solution (a) Energy of the incident photon, E = hν = hc/λ E = (6.63×10–34J s) (3×108 m/s)/λ = 1.989 × 10–25 J m λ (i) For violet light, λ1 = 390 nm (lower wavelength end) 1.989 × 10–25 J m Incident photon energy, E1 = 390×10–9 m = 5.10 × 10–19J = 5.10 × 10–19 J 1.6×10–19 J/eV = 3.19 eV (ii) For yellow-green light, λ2 = 550 nm (average wavelength) 1.989 × 10–25 J m Incident photon energy, E2 = 550×10–9 m = 3.62×10–19 J = 2.26 eV (iii) For red light, λ3 = 760 nm (higher wavelength end) 1.989 × 10–25 J m Incident photon energy, E3 = 760×10–9 m = 2.62×10–19 J = 1.64 eV (b) For a photoelectric device to operate, we require incident light energy E to be equal to or greater than the work function φ0 of the material. EXAMPLE 11.3 Thus, the photoelectric device will operate with violet light (with E = 3.19 eV) photosensitive material Na (with wφ0ill= 2.75 eV), K (with φ0 = 2.30 eV ) and Cs (with φ0 = 2.14 eV). It also operate with yellow-green light (with E = 2.26 eV) for Cs (with φ0 = 2.14 eV) only. However, it will not operate with red light (with E = 1.64 eV) for any of these photosensitive materials. 397 2020-21

Physics 11.8 WAVE NATURE OF MATTER The dual (wave-particle) nature of light (electromagnetic radiation, in general) comes out clearly from what we have learnt in this and the preceding chapters. The wave nature of light shows up in the phenomena of interference, diffraction and polarisation. On the other hand, in photoelectric effect and Compton effect which involve energy and momentum transfer, radiation behaves as if it is made up of a bunch of particles – the photons. Whether a particle or wave description is best suited for understanding an experiment depends on the nature of the experiment. For example, in the familiar phenomenon of seeing an object by our eye, both descriptions are important. The gathering and focussing mechanism of light by the eye-lens is well described in the wave picture. But its absorption by the rods and cones (of the retina) requires the photon picture of light. A natural question arises: If radiation has a dual (wave-particle) nature, might not the particles of nature (the electrons, protons, etc.) also exhibit wave-like character? In 1924, the French physicist Louis Victor de Broglie (pronounced as de Broy) (1892-1987) put forward the bold hypothesis that moving particles of matter should display wave-like properties under suitable conditions. He reasoned that nature was symmetrical and that the two basic physical entities – matter and energy, must have symmetrical character. If radiation shows dual aspects, so should matter. De Broglie proposed that the wave length λ associated with a particle of momentum p is given as λ=h= h (11.5) p mv where m is the mass of the particle and v its speed. Equation (11.5) is known as the de Broglie relation and the wavelength λ of the matter wave is called de Broglie wavelength. The dual aspect of matter is evident in the de Broglie relation. On the left hand side of Eq. (11.5), λ is the attribute of a wave while on the right hand side the momentum p is a typical attribute of a particle. Planck’s constant h relates the two attributes. Equation (11.5) for a material particle is basically a hypothesis whose validity can be tested only by experiment. However, it is interesting to see that it is satisfied also by a photon. For a photon, as we have seen, p = hν /c (11.6) Therefore, h =c =λ (11.7) pν That is, the de Broglie wavelength of a photon given by Eq. (11.5) equals the wavelength of electromagnetic radiation of which the photon is a quantum of energy and momentum. Clearly, from Eq. (11.5 ), λ is smaller for a heavier particle ( large m ) or more energetic particle (large v). For example, the de Broglie wavelength of a ball of mass 0.12 kg moving with a speed of 20 m s–1 is easily 398 calculated: 2020-21

Dual Nature of Radiation and Matter PHOTOCELL A photocell is a technological application of the photoelectric effect. It is a device whose electrical properties are affected by light. It is also sometimes called an electric eye. A photocell consists of a semi-cylindrical photo-sensitive metal plate C (emitter) and a wire loop A (collector) supported in an evacuated glass or quartz bulb. It is connected to the external circuit having a high-tension battery B and microammeter (µA) as shown in the Figure. Sometimes, instead of the plate C, a thin layer of photosensitive material is pasted on the inside of the bulb. A part of the bulb is left clean for the light to enter it. When light of suitable wavelength falls on the emitter C, photoelectrons are emitted. These photoelectrons are drawn to the collector A. Photocurrent of the order of a few microampere can be normally obtained from a photo cell. A photocell converts a change in intensity of illumination into a change in photocurrent. This current can be used to operate control systems and in light measuring devices. A photocell of lead sulphide sensitive to infrared radiation is used in electronic ignition circuits. In scientific work, photo cells are used whenever it is necessary to measure the intensity of light. Light meters in photographic cameras make use of photo cells to measure the intensity of incident light. The photocells, inserted in the door light electric circuit, are used as automatic door opener. A person approaching a doorway may interrupt a light beam which is incident on a photocell. The abrupt change in photocurrent may be used to start a motor which opens the A photo cell door or rings an alarm. They are used in the control of a counting device which records every interruption of the light beam caused by a person or object passing across the beam. So photocells help count the persons entering an auditorium, provided they enter the hall one by one. They are used for detection of traffic law defaulters: an alarm may be sounded whenever a beam of (invisible) radiation is intercepted. In burglar alarm, (invisible) ultraviolet light is continuously made to fall on a photocell installed at the doorway. A person entering the door interrupts the beam falling on the photocell. The abrupt change in photocurrent is used to start an electric bell ringing. In fire alarm, a number of photocells are installed at suitable places in a building. In the event of breaking out of fire, light radiations fall upon the photocell. This completes the electric circuit through an electric bell or a siren which starts operating as a warning signal. Photocells are used in the reproduction of sound in motion pictures and in the television camera for scanning and telecasting scenes. They are used in industries for detecting minor flaws or holes in metal sheets. p = m v = 0.12 kg × 20 m s–1 = 2.40 kg m s–1 λ= h = 6.63 × 10−34 J s = 2.76 × 10–34 m 399 p 2.40 kg m s−1 2020-21

Physics LOUIS VICTOR DE BROGLIE (1892 – 1987) This wavelength is so small that it is beyond any measurement. This is the reason why macroscopic objects in our daily life do not show wave-like properties. On the other hand, in the sub-atomic domain, the wave character of particles is significant and measurable. Consider an electron (mass m, charge e) accelerated from rest through a potential V. The kinetic energy K of the electron equals the work done (eV ) on it by the electric field: K=eV (11.8) Now, K = 1 m v2 = p2 , so that 2 2m Louis Victor de Broglie p= 2 m K = 2 m eV (11.9) (1892 – 1987) French The de Broglie wavelength λ of the electron is then physicist who put forth revolutionary idea of wave λ= h= h K = h (11.10) p 2m 2 m eV nature of matter. This idea was developed by Erwin Schródinger into a full- Substituting the numerical values of h, m, e, fledged theory of quantum we get mechanics commonly λ = 1.227 nm (11.11) known as wave mechanics. V In 1929, he was awarded the where V is the magnitude of accelerating potential in Nobel Prize in Physics for his volts. For a 120 V accelerating potential, Eq. (11.11) gives discovery of the wave nature of electrons. λ = 0.112 nm. This wavelength is of the same order as the spacing between the atomic planes in crystals. This suggests that matter waves associated with an electron could be verified by crystal diffraction experiments analogous to X-ray diffraction. We describe the experimental verification of the de Broglie hypothesis in the next section. In 1929, de Broglie was awarded the Nobel Prize in Physics for his discovery of the wave nature of electrons. The matter–wave picture elegantly incorporated the Heisenberg’s uncertainty principle. According to the principle, it is not possible to measure both the position and momentum of an electron (or any other particle) at the same time exactly. There is always some uncertainty (∆ x ) in the specification of position and some uncertainty (∆p ) in the specification of momentum. The product of ∆x and ∆p is of the order of ħ* (with ħ = h/2π), i.e., ∆x ∆p ≈ ħ (11.12) Equation (11.12) allows the possibility that ∆x is zero; but then ∆p must be infinite in order that the product is non-zero. Similarly, if ∆p is zero, ∆x must be infinite. Ordinarily, both ∆x and ∆p are non-zero such that their product is of the order of ħ. Now, if an electron has a definite momentum p, (i.e.∆p = 0), by the de Broglie relation, it has a definite wavelength λ. A wave of definite (single) 400 * A more rigorous treatment gives ∆x ∆p ≥ ħ/2. 2020-21

Dual Nature of Radiation and Matter wavelength extends all over space. By Born’s FIGURE 11.6 (a) The wave packet description of probability interpretation this means that the an electron. The wave packet corresponds to a spread of wavelength around some central electron is not localised in any finite region of space. That is, its position uncertainty is infinite wavelength (and hence by de Broglie relation, a spread in momentum). Consequently, it is (∆x → ∞), which is consistent with the uncertainty principle. associated with an uncertainty in position (∆x) and an uncertainty in momentum (∆p). In general, the matter wave associated with the electron is not extended all over space. It is (b) The matter wave corresponding to a definite momentum of an electron a wave packet extending over some finite region of space. In that case ∆x is not infinite but has extends all over space. In this case, ∆p = 0 and ∆ x → ∞. some finite value depending on the extension of the wave packet. Also, you must appreciate that a wave packet of finite extension does not have a single wavelength. It is built up of wavelengths spread around some central wavelength. By de Broglie’s relation, then, the momentum of the electron will also have a spread – an uncertainty ∆p. This is as expected from the uncertainty principle. It can be shown that the wave packet description together with de Broglie relation and Born’s probability interpretation reproduce the Heisenberg’s uncertainty principle exactly. In Chapter 12, the de Broglie relation will be seen to justify Bohr’s postulate on quantisation of angular momentum of electron in an atom. Figure 11.6 shows a schematic diagram of (a) a localised wave packet, and (b) an extended wave with fixed wavelength. Example 11.4 What is the de Broglie wavelength associated with (a) an electron moving with a speed of 5.4×106 m/s, and (b) a ball of mass 150 g travelling at 30.0 m/s? Solution (a) For the electron: Mass m = 9.11×10–31 kg, speed v = 5.4×106 m/s. Then, momentum p = m v = 9.11×10–31 (kg) × 5.4 × 106 (m/s) p = 4.92 × 10–24 kg m/s de Broglie wavelength, λ = h/p = 6.63 ×10–34 J s 4.92 ×10–24 kg m/s λ = 0.135 nm EXAMPLE 11.4 401 (b) For the ball: Mass m ’ = 0.150 kg, speed v ’ = 30.0 m/s. Then momentum p ’ = m’ v ’ = 0.150 (kg) × 30.0 (m/s) p ’= 4.50 kg m/s de Broglie wavelength λ’ = h/p’. 2020-21

Physics EXAMPLE 11.4 6.63 ×10–34 J s = 4.50 × kg m/s λ ’ = 1.47 ×10–34 m The de Broglie wavelength of electron is comparable with X-ray wavelengths. However, for the ball it is about 10–19 times the size of the proton, quite beyond experimental measurement. Example 11.5 An electron, an α-particle, and a proton have the same kinetic energy. Which of these particles has the shortest de Broglie wavelength? Solution For a particle, de Broglie wavelength, λ = h/p Kinetic energy, K = p2/2m EXAMPLE 11.5 Then, λ = h / 2mK For the same kinetic energy K, the de Broglie wavelength associated with the particle is inversely proportional to the square root of their ( )masses. A proton1 H is 1836 times massive than an electron and 1 ( )an α-particle4Hefour times that of a proton. 2 Hence, α – particle has the shortest de Broglie wavelength. PROBABILITY INTERPRETATION TO MATTER WAVES It is worth pausing here to reflect on just what a matter wave associated with a particle, say, an electron, means. Actually, a truly satisfactory physical understanding of the dual nature of matter and radiation has not emerged so far. The great founders of quantum mechanics (Niels Bohr, Albert Einstein, and many others) struggled with this and related concepts for long. Still the deep physical interpretation of quantum mechanics continues to be an area of active research. Despite this, the concept of matter wave has been mathematically introduced in modern quantum mechanics with great success. An important milestone in this connection was when Max Born (1882- 1970) suggested a probability interpretation to the matter wave amplitude. According to this, the intensity (square of the amplitude) of the matter wave at a point determines the probability density of the particle at that point. Probability density means probability per unit volume. Thus, if A is the amplitude of the wave at a point, |A|2 ∆V is the probability of the particle being found in a small volume ∆V around that point. Thus, if the intensity of matter wave is large in a certain region, there is a greater probability of the particle being found there than where the intensity is small. 402 EXAMPLE 11.6 Example 11.6 A particle is moving three times as fast as an electron. The ratio of the de Broglie wavelength of the particle to that of the electron is 1.813 × 10–4. Calculate the particle’s mass and identify the particle. Solution de Broglie wavelength of a moving particle, having mass m and velocity v: 2020-21

Dual Nature of Radiation and Matter λ=h = h p mv Mass, m = h/λv For an electron, mass me = h/λe ve Now, =we1.h81av3e×v1/0ve– = 3 and λ/λe 4 Then, mass of the particle, m = me  λe   ve  EXAMPLE 11.6  λ   v  m = (9.11×10–31 kg) × (1/3) × (1/1.813 × 10–4) m = 1.675 × 10–27 kg. Thus, the particle, with this mass could be a proton or a neutron. Example 11.7 What is the de Broglie wavelength associated with an electron, accelerated through a potential differnece of 100 volts? Solution Accelerating potential V = 100 V. The de Broglie wavelength EXAMPLE 11.7 λ is λ = h /p = 1.227 nm V λ 1.227 nm = 0.123 nm = 100 The de Broglie wavelength associated with an electron in this case is of the order of X-ray wavelengths. 11.9 DAVISSON AND GERMER EXPERIMENT The wave nature of electrons was first experimentally verified by C.J. Davisson and L.H. Germer in 1927 and independently by G.P. Thomson, in 1928, who observed diffraction effects with beams of electrons scattered by crystals. Davisson and Thomson shared the Nobel Prize in 1937 for their experimental discovery of diffraction of electrons by crystals. The experimental arrange- ment used by Davisson and Germer is schematically shown in Fig. 11.7. It consists of an electron gun which comprises of a tungsten filament F, coated with barium oxide and heated by a low voltage power supply (L.T. or battery). Electrons emitted by the filament are FIGURE 11.7 Davisson-Germer electron 403 accelerated to a desired velocity diffraction arrangement. 2020-21

Physics Development of electron microscope by applying suitable potential/voltage from a high voltage power supply http://www.nobelprize.org/nobel_prizes/physics/laureates/1986/presentation-speech.html (H.T. or battery). They are made to pass through a cylinder with fine holes along its axis, producing a fine collimated beam. The beam is made to fall on the surface of a nickel crystal. The electrons are scattered in all directions by the atoms of the crystal. The intensity of the electron beam, scattered in a given direction, is measured by the electron detector (collector). The detector can be moved on a circular scale and is connected to a sensitive galvanometer, which records the current. The deflection of the galvanometer is proportional to the intensity of the electron beam entering the collector. The apparatus is enclosed in an evacuated chamber. By moving the detector on the circular scale at different positions, the intensity of the scattered electron beam is measured for different values of angle of scattering θ which is the angle between the incident and the scattered electron beams. The variation of the intensity (I ) of the scattered electrons with the angle of scattering θ is obtained for different accelerating voltages. The experiment was performed by varying the accelarating voltage from 44 V to 68 V. It was noticed that a strong peak appeared in the intensity (I ) of the scattered electron for an accelarating voltage of 54V at a scattering angle θ = 50° The appearance of the peak in a particular direction is due to the constructive interference of electrons scattered from different layers of the regularly spaced atoms of the crystals. From the electron diffraction measurements, the wavelength of matter waves was found to be 0.165 nm. The de Broglie wavelength λ associated with electrons, using Eq. (11.11), for V = 54 V is given by λ= h /p 1 227 nm = V λ 1 227 nm = 0.167 nm = 54 Thus, there is an excellent agreement between the theoretical value and the experimentally obtained value of de Broglie wavelength. Davisson- Germer experiment thus strikingly confirms the wave nature of electrons and the de Broglie relation. More recently, in 1989, the wave nature of a beam of electrons was experimentally demonstrated in a double-slit experiment, similar to that used for the wave nature of light. Also, in an experiment in 1994, interference fringes were obtained with the beams of iodine molecules, which are about a million times more massive than electrons. The de Broglie hypothesis has been basic to the development of modern quantum mechanics. It has also led to the field of electron optics. The wave properties of electrons have been utilised in the design of electron microscope which is a great improvement, with higher resolution, over the optical microscope. 404 2020-21

Dual Nature of Radiation and Matter SUMMARY 1. The minimum energy needed by an electron to come out from a metal surface is called the work function of the metal. Energy (greater than the work function (φο) required for electron emission from the metal surface can be supplied by suitably heating or applying strong electric field or irradiating it by light of suitable frequency. 2. Photoelectric effect is the phenomenon of emission of electrons by metals when illuminated by light of suitable frequency. Certain metals respond to ultraviolet light while others are sensitive even to the visible light. Photoelectric effect involves conversion of light energy into electrical energy. It follows the law of conservation of energy. The photoelectric emission is an instantaneous process and possesses certain special features. 3. Photoelectric current depends on (i) the intensity of incident light, (ii) the potential difference applied between the two electrodes, and (iii) the nature of the emitter material. 4. The stopping potential (Vo) depends on (i) the frequency of incident light, and (ii) the nature of the emitter material. For a given frequency of incident light, it is independent of its intensity. The stopping potential is directly related to the maximum kinetic energy of electrons emitted: e V0 = (1/2) m v2 = Kmax. max 5. Below a certain frequency (threshold frequency) ν 0, characteristic of the metal, no photoelectric emission takes place, no matter how large the intensity may be. 6. The classical wave theory could not explain the main features of photoelectric effect. Its picture of continuous absorption of energy from radiation could not explain the independence of Kmax on intensity, the existence of νo and the instantaneous nature of the process. Einstein explained these features on the basis of photon picture of light. According to this, light is composed of discrete packets of energy called quanta or photons. Each photon carries an energy E (= h ν) and momentum p (= h/λ), which depend on the frequency (ν ) of incident light and not on its intensity. Photoelectric emission from the metal surface occurs due to absorption of a photon by an electron. 7. Einstein’s photoelectric equation is in accordance with the energy conservation law as applied to the photon absorption by an electron in the metal. The maximum kinetic energy (1/2)m v2 is equal to max the photon energy (hν ) minus the work function φ0 (= hν0) of the target metal: 1 m v2 = V0 e = hν – φ0 = h (ν – ν0 ) max 2 This photoelectric equation explains all the features of the photoelectric 405 effect. Millikan’s first precise measurements confirmed the Einstein’s photoelectric equation and obtained an accurate value of Planck’s constant h . This led to the acceptance of particle or photon description (nature) of electromagnetic radiation, introduced by Einstein. 8. Radiation has dual nature: wave and particle. The nature of experiment determines whether a wave or particle description is best suited for understanding the experimental result. Reasoning that radiation and matter should be symmetrical in nature, Louis Victor de Broglie 2020-21

Physics attributed a wave-like character to matter (material particles). The waves associated with the moving material particles are called matter waves or de Broglie waves. 9. The de Broglie wavelength (λ) associated with a moving particle is related to its momentum p as: λ = h/p. The dualism of matter is inherent in the de Broglie relation which contains a wave concept (λ) and a particle concept (p). The de Broglie wavelength is independent of the charge and nature of the material particle. It is significantly measurable (of the order of the atomic-planes spacing in crystals) only in case of sub-atomic particles like electrons, protons, etc. (due to smallness of their masses and hence, momenta). However, it is indeed very small, quite beyond measurement, in case of macroscopic objects, commonly encountered in everyday life. 10. Electron diffraction experiments by Davisson and Germer, and by G. P. Thomson, as well as many later experiments, have verified and confirmed the wave-nature of electrons. The de Broglie hypothesis of matter waves supports the Bohr ’s concept of stationary orbits. Physical Symbol Dimensions Unit Remarks Quantity E = hν Planck’s h [ML2 T –1] Js e V0= Kmax constant Kmax = E –φ0 V0 [ML2 T –3A–1] V ν0 = φ0 /h Stopping φ0 potential ν0 [ML2 T –2] J; eV λ= h/p λ Work [T –1] Hz function [L] m Threshold frequency de Broglie wavelength 406 POINTS TO PONDER 1. Free electrons in a metal are free in the sense that they move inside the metal in a constant potential (This is only an approximation). They are not free to move out of the metal. They need additional energy to get out of the metal. 2. Free electrons in a metal do not all have the same energy. Like molecules in a gas jar, the electrons have a certain energy distribution at a given temperature. This distribution is different from the usual Maxwell’s distribution that you have learnt in the study of kinetic theory of gases. You will learn about it in later courses, but the difference has to do with the fact that electrons obey Pauli’s exclusion principle. 3. Because of the energy distribution of free electrons in a metal, the energy required by an electron to come out of the metal is different for different electrons. Electrons with higher energy require less additional energy to come out of the metal than those with lower energies. Work function is the least energy required by an electron to come out of the metal. 2020-21

Dual Nature of Radiation and Matter 4. Observations on photoelectric effect imply that in the event of matter- light interaction, absorption of energy takes place in discrete units of hν. This is not quite the same as saying that light consists of particles, each of energy hν. 5. Observations on the stopping potential (its independence of intensity and dependence on frequency) are the crucial discriminator between the wave-picture and photon-picture of photoelectric effect. 6. The wavelength of a matter wave given by λ = h has physical p significance; its phase velocity vp has no physical significance. However, the group velocity of the matter wave is physically meaningful and equals the velocity of the particle. EXERCISES 11.1 Find the 11.2 (a) maximum frequency, and 11.3 (b) minimum wavelength of X-rays produced by 30 kV electrons. 11.4 The work function of caesium metal is 2.14 eV. When light of 11.5 frequency 6 ×1014Hz is incident on the metal surface, photoemission 11.6 of electrons occurs. What is the 11.7 (a) maximum kinetic energy of the emitted electrons, 407 (b) Stopping potential, and (c) maximum speed of the emitted photoelectrons? The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted? Monochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW. (a) Find the energy and momentum of each photon in the light beam, (b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area ), and (c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon? The energy flux of sunlight reaching the surface of the earth is 1.388 × 103 W/m2. How many photons (nearly) per square metre are incident on the Earth per second ? Assume that the photons in the sunlight have an average wavelength of 550 nm. In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 × 10–15 V s. Calculate the value of Planck’s constant. A 100W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm. (a) What is the energy per photon associated 2020-21

Physics with the sodium light? (b) At what rate are the photons delivered to the sphere? 11.8 The threshold frequency for a certain metal is 3.3 × 1014 Hz. If light of frequency 8.2 × 1014 Hz is incident on the metal, predict the cut- off voltage for the photoelectric emission. 11.9 The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm? 11.10 Light of frequency 7.21 × 1014 Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 105 m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons? 11.11 Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-of f) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made. 11.12 Calculate the (a) momentum, and (b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V. 11.13 What is the (a) momentum, (b) speed, and (c) de Broglie wavelength of an electron with kinetic energy of 120 eV. 11.14 The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which (a) an electron, and (b) a neutron, would have the same de Broglie wavelength. 11.15 What is the de Broglie wavelength of (a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s, (b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and (c) a dust particle of mass 1.0 × 10–9 kg drifting with a speed of 2.2 m/s ? 11.16 An electron and a photon each have a wavelength of 1.00 nm. Find (a) their momenta, (b) the energy of the photon, and (c) the kinetic energy of electron. 11.17 (a) For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 × 10–10 m? (b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of (3/2) k T at 300 K. 11.18 Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon). 408 11.19 What is the de Broglie wavelength of a nitrogen molecule in air at 300 K ? Assume that the molecule is moving with the root-mean- square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u) 2020-21

Dual Nature of Radiation and Matter ADDITIONAL EXERCISES 409 11.20 (a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter. Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., its e/m is given to be 1.76 × 1011 C kg–1. (b) Use the same formula you employ in (a) to obtain electron speed for an collector potential of 10 MV. Do you see what is wrong ? In what way is the formula to be modified? 11.21 (a) A monoenergetic electron beam with electron speed of 5.20 × 106 m s–1 is subject to a magnetic field of 1.30 × 10–4 T normal to the beam velocity. What is the radius of the circle traced by the beam, given e/m for electron equals 1.76 × 1011C kg–1. (b) Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam? If not, in what way is it modified ? [Note: Exercises 11.20(b) and 11.21(b) take you to relativistic mechanics which is beyond the scope of this book. They have been inserted here simply to emphasise the point that the formulas you use in part (a) of the exercises are not valid at very high speeds or energies. See answers at the end to know what ‘very high speed or energy’ means.] 11.22 An electron gun with its collector at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (∼10–2 mm of Hg). A magnetic field of 2.83 × 10–4 T curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the ‘fine beam tube’ method.) Determine e/m from the data. 11.23 (a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 Å. What is the maximum energy of a photon in the radiation ? (b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube ? 11.24 In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as annihilation of an electron-positron pair of total energy 10.2 BeV into two γ-rays of equal energy. What is the wavelength associated with each γ-ray? (1BeV = 109 eV) 11.25 Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never ‘count photons’, even in barely detectable light. (a) The number of photons emitted per second by a Medium wave transmitter of 10 kW power, emitting radiowaves of wavelength 500 m. (b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we 2020-21

410 Physics humans can perceive (∼10–10 W m–2). Take the area of the pupil to be about 0.4 cm2, and the average frequency of white light to be about 6 × 1014 Hz. 11.26 Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is –1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity (∼105 W m–2) red light of wavelength 6328 Å produced by a He-Ne laser ? 11.27 Monochromatic radiation of wavelength 640.2 nm (1nm = 10–9 m) from a neon lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photo-cell. Predict the new stopping voltage. 11.28 A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission, since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used: λ1 = 3650 Å, λ2= 4047 Å, λ3= 4358 Å, λ4= 5461 Å, λ5= 6907 Å, The stopping voltages, respectively, were measured to be: V01 = 1.28 V, V02 = 0.95 V, V03 = 0.74 V, V04 = 0.16 V, V05 = 0 V Determine the value of Planck’s constant h, the threshold frequency and work function for the material. [Note: You will notice that to get h from the data, you will need to know e (which you can take to be 1.6 × 10–19 C). Experiments of this kind on Na, Li, K, etc. were performed by Millikan, who, using his own value of e (from the oil-drop experiment) confirmed Einstein’s photoelectric equation and at the same time gave an independent estimate of the value of h.] 11.29 The work function for the following metals is given: Na: 2.75 eV; K: 2.30 eV; Mo: 4.17 eV; Ni: 5.15 eV. Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 Å from a He-Cd laser placed 1 m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away ? 11.30 Light of intensity 10–5 W m–2 falls on a sodium photo-cell of surface area 2 cm2. Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer ? 11.31 Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1 Å, which is of the order of inter-atomic spacing in the lattice) (me=9.11 × 10–31 kg). 11.32 (a) Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have seen in Exercise 11.31, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable ? Explain. (mn = 1.675 × 10–27 kg) 2020-21

Dual Nature of Radiation and Matter (b) Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (27 °C). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments. 11.33 An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light? 11.34 The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale of 10–15 m or less. This structure was first probed in early 1970’s using high energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams. (Rest mass energy of electron = 0.511 MeV.) 11.35 Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27 °C) and 1 atm pressure; and compare it with the mean separation between two atoms under these conditions. 11.36 Compute the typical de Broglie wavelength of an electron in a metal at 27 °C and compare it with the mean separation between two electrons in a metal which is given to be about 2 × 10–10 m. [Note: Exercises 11.35 and 11.36 reveal that while the wave-packets associated with gaseous molecules under ordinary conditions are non-overlapping, the electron wave-packets in a metal strongly overlap with one another. This suggests that whereas molecules in an ordinary gas can be distinguished apart, electrons in a metal cannot be distintguished apart from one another. This indistinguishibility has many fundamental implications which you will explore in more advanced Physics courses.] 11.37 Answer the following questions: (a) Quarks inside protons and neutrons are thought to carry fractional charges [(+2/3)e ; (–1/3)e]. Why do they not show up in Millikan’s oil-drop experiment? (b) What is so special about the combination e/m? Why do we not simply talk of e and m separately? (c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures? (d) Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic? Why is there an energy distribution of photoelectrons? (e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations: E = h ν, p = h λ But while the value of λ is physically significant, the value of ν 411 (and therefore, the value of the phase speed ν λ) has no physical significance. Why? 2020-21

Physics APPENDIX 11.1 The history of wave-particle flip-flop What is light? This question has haunted mankind for a long time. But systematic experiments were done by scientists since the dawn of the scientific and industrial era, about four centuries ago. Around the same time, theoretical models about what light is made of were developed. While building a model in any branch of science, it is essential to see that it is able to explain all the experimental observations existing at that time. It is therefore appropriate to summarize some observations about light that were known in the seventeenth century. The properties of light known at that time included (a) rectilinear propagation of light, (b) reflection from plane and curved surfaces, (c) refraction at the boundary of two media, (d) dispersion into various colours, (e) high speed. Appropriate laws were formulated for the first four phenomena. For example, Snell formulated his laws of refraction in 1621. Several scientists right from the days of Galileo had tried to measure the speed of light. But they had not been able to do so. They had only concluded that it was higher than the limit of their measurement. Two models of light were also proposed in the seventeenth century. Descartes, in early decades of seventeenth century, proposed that light consists of particles, while Huygens, around 1650-60, proposed that light consists of waves. Descartes′ proposal was merely a philosophical model, devoid of any experiments or scientific arguments. Newton soon after, around 1660-70, extended Descartes′ particle model, known as corpuscular theory, built it up as a scientific theory, and explained various known properties with it. These models, light as waves and as particles, in a sense, are quite opposite of each other. But both models could explain all the known properties of light. There was nothing to choose between them. The history of the development of these models over the next few centuries is interesting. Bartholinus, in 1669, discovered double refraction of light in some crystals, and Huygens, in 1678, was quick to explain it on the basis of his wave theory of light. In spite of this, for over one hundred years, Newton’s particle model was firmly believed and preferred over the wave model. This was partly because of its simplicity and partly because of Newton’s influence on contemporary physics. Then in 1801, Young performed his double-slit experiment and observed interference fringes. This phenomenon could be explained only by wave theory. It was realized that diffraction was also another phenomenon which could be explained only by wave theory. In fact, it was a natural consequence of Huygens idea of secondary wavelets emanating from every point in the path of light. These experiments could not be explained by assuming that light consists of particles. Another phenomenon of polarisation was discovered around 1810, and this too could be naturally explained by the wave theory. Thus wave theory of Huygens came to the forefront and Newton’s particle theory went into the background. This situation again continued for almost a century. Better experiments were performed in the nineteenth century to determine the speed of light. With more accurate experiments, a value of 3×108 m/s for speed of light in vacuum was arrived at. Around 1860, Maxwell proposed his equations of electromagnetism and it was realized that all electromagnetic phenomena known at that time could be explained by Maxwell’s four equations. Soon Maxwell showed that electric and magnetic fields could propagate through empty space (vacuum) in the form of electromagnetic waves. He calculated the speed of these waves and arrived at a theoretical value of 2.998×108 m/s. The close agreement of this value with the experimental value suggested that light consists of electromagnetic waves. In 1887 Hertz demonstrated the generation and detection of such waves. This established the wave theory of light on a firm footing. We might say that while eighteenth century belonged to the particle model, the nineteenth century belonged to the wave model of light. Vast amounts of experiments were done during the period 1850-1900 on heat and related phenomena, an altogether different area of physics. Theories and models like kinetic theory and thermodynamics were developed which quite successfully explained the various phenomena, except one. 412 2020-21

Dual Nature of Radiation and Matter Every body at any temperature emits radiation of all wavelengths. It also absorbs radiation falling on it. A body which absorbs all the radiation falling on it is called a black body. It is an ideal concept in physics, like concepts of a point mass or uniform motion. A graph of the intensity of radiation emitted by a black body versus wavelength is called the black body spectrum. No theory in those days could explain the complete black body spectrum! In 1900, Planck hit upon a novel idea. If we assume, he said, that radiation is emitted in packets of energy instead of continuously as in a wave, then we can explain the black body spectrum. Planck himself regarded these quanta, or packets, as a property of emission and absorption, rather than that of light. He derived a formula which agreed with the entire spectrum. This was a confusing mixture of wave and particle pictures – radiation is emitted as a particle, it travels as a wave, and is again absorbed as a particle! Moreover, this put physicists in a dilemma. Should we again accept the particle picture of light just to explain one phenomenon? Then what happens to the phenomena of interference and diffraction which cannot be explained by the particle model? But soon in 1905, Einstein explained the photoelectric effect by assuming the particle picture of light. In 1907, Debye explained the low temperature specific heats of solids by using the particle picture for lattice vibrations in a crystalline solid. Both these phenomena belonging to widely diverse areas of physics could be explained only by the particle model and not by the wave model. In 1923, Compton’s x-ray scattering experiments from atoms also went in favour of the particle picture. This increased the dilemma further. Thus by 1923, physicists faced with the following situation. (a) There were some phenomena like rectilinear propagation, reflection, refraction, which could be explained by either particle model or by wave model. (b) There were some phenomena such as diffraction and interference which could be explained only by the wave model but not by the particle model. (c) There were some phenomena such as black body radiation, photoelectric effect, and Compton scattering which could be explained only by the particle model but not by the wave model. Somebody in those days aptly remarked that light behaves as a particle on Mondays, Wednesdays and Fridays, and as a wave on Tuesdays, Thursdays and Saturdays, and we don’t talk of light on Sundays! In 1924, de Broglie proposed his theory of wave-particle duality in which he said that not only photons of light but also ‘particles’ of matter such as electrons and atoms possess a dual character, sometimes behaving like a particle and sometimes as a wave. He gave a formula connecting their mass, velocity, momentum (particle characteristics), with their wavelength and frequency (wave characteristics)! In 1927 Thomson, and Davisson and Germer, in separate experiments, showed that electrons did behave like waves with a wavelength which agreed with that given by de Broglie’s formula. Their experiment was on diffraction of electrons through crystalline solids, in which the regular arrangement of atoms acted like a grating. Very soon, diffraction experiments with other ‘particles’ such as neutrons and protons were performed and these too confirmed with de Broglie’s formula. This confirmed wave-particle duality as an established principle of physics. Here was a principle, physicists thought, which explained all the phenomena mentioned above not only for light but also for the so-called particles. But there was no basic theoretical foundation for wave-particle duality. De Broglie’s proposal was merely a qualitative argument based on symmetry of nature. Wave-particle duality was at best a principle, not an outcome of a sound fundamental theory. It is true that all experiments whatever agreed with de Broglie formula. But physics does not work that way. On the one hand, it needs experimental confirmation, while on the other hand, it also needs sound theoretical basis for the models proposed. This was developed over the next two decades. Dirac developed his theory of radiation in about 1928, and Heisenberg and Pauli gave it a firm footing by 1930. Tomonaga, Schwinger, and Feynman, in late 1940s, produced further refinements and cleared the theory of inconsistencies which were noticed. All these theories mainly put wave-particle duality on a theoretical footing. Although the story continues, it grows more and more complex and beyond the scope of this note. But we have here the essential structure of what happened, and let us be satisfied with it at the moment. Now it is regarded as a natural consequence of present theories of physics that electromagnetic radiation as well as particles of matter exhibit both wave and particle properties in different experiments, and sometimes even in the different parts of the same experiment. 413 2020-21

Physics Chapter Twelve ATOMS 12.1 INTRODUCTION By the nineteenth century, enough evidence had accumulated in favour of atomic hypothesis of matter. In 1897, the experiments on electric discharge through gases carried out by the English physicist J. J. Thomson (1856 – 1940) revealed that atoms of different elements contain negatively charged constituents (electrons) that are identical for all atoms. However, atoms on a whole are electrically neutral. Therefore, an atom must also contain some positive charge to neutralise the negative charge of the electrons. But what is the arrangement of the positive charge and the electrons inside the atom? In other words, what is the structure of an atom? The first model of atom was proposed by J. J. Thomson in 1898. According to this model, the positive charge of the atom is uniformly distributed throughout the volume of the atom and the negatively charged electrons are embedded in it like seeds in a watermelon. This model was picturesquely called plum pudding model of the atom. However subsequent studies on atoms, as described in this chapter, showed that the distribution of the electrons and positive charges are very different from that proposed in this model. We know that condensed matter (solids and liquids) and dense gases at all temperatures emit electromagnetic radiation in which a continuous 414 distribution of several wavelengths is present, though with different intensities. This radiation is considered to be due to oscillations of atoms 2020-21

Atoms and molecules, governed by the interaction of each atom or molecule with its neighbours. In contrast, light emitted from rarefied gases heated in a flame, or excited electrically in a glow tube such as the familiar neon sign or mercury vapour light has only certain discrete wavelengths. The spectrum appears as a series of bright lines. In such gases, the average spacing between atoms is large. Hence, the radiation emitted can be considered due to individual atoms rather than because of interactions between atoms or molecules. In the early nineteenth century it was also established that each element is associated with a characteristic spectrum of radiation, for example, hydrogen always gives a set of lines with fixed relative position between the lines. Ernst Rutherford (1871 – ERNST RUTHERFORD (1871 – 1937) This fact suggested an intimate relationship between the 1937) New Zealand born, internal structure of an atom and the spectrum of British physicist who did radiation emitted by it. In 1885, Johann Jakob Balmer (1825 – 1898) obtained a simple empirical formula which pioneering work on gave the wavelengths of a group of lines emitted by atomic radioactive radiation. He hydrogen. Since hydrogen is simplest of the elements discovered alpha-rays and known, we shall consider its spectrum in detail in this chapter. beta-rays. Along with Federick Soddy, he created the modern theory of Ernst Rutherford (1871–1937), a former research radioactivity. He studied student of J. J. Thomson, was engaged in experiments on the ‘emanation’ of thorium and discovered a new noble α-particles emitted by some radioactive elements. In 1906, he proposed a classic experiment of scattering of these gas, an isotope of radon, α-particles by atoms to investigate the atomic structure. now known as thoron. By This experiment was later performed around 1911 by Hans scattering alpha-rays from Geiger (1882–1945) and Ernst Marsden (1889–1970, who was 20 year-old student and had not yet earned his the metal foils, he bachelor’s degree). The details are discussed in Section discovered the atomic 12.2. The explanation of the results led to the birth of nucleus and proposed the Rutherford’s planetary model of atom (also called the nuclear model of the atom). According to this the entire plenatery model of the positive charge and most of the mass of the atom is atom. He also estimated the approximate size of the nucleus. concentrated in a small volume called the nucleus with electrons revolving around the nucleus just as planets revolve around the sun. Rutherford’s nuclear model was a major step towards how we see the atom today. However, it could not explain why atoms emit light of only discrete wavelengths. How could an atom as simple as hydrogen, consisting of a single electron and a single proton, emit a complex spectrum of specific wavelengths? In the classical picture of an atom, the electron revolves round the nucleus much like the way a planet revolves round the sun. However, we shall see that there are some serious difficulties in accepting such a model. 12.2 ALPHA-PARTICLE SCATTERING AND 415 RUTHERFORD’S NUCLEAR MODEL OF ATOM At the suggestion of Ernst Rutherford, in 1911, H. Geiger and E. Marsden performed some experiments. In one of their experiments, as shown in 2020-21

Physics Fig. 12.1, they directed a beam of 5.5 MeV α-particles emitted from a 214 Bi radioactive source at a thin metal 83 foil made of gold. Figure 12.2 shows a schematic diagram of this experiment. Alpha-particles emitted by a 214 Bi 83 radioactive source were collimated into a narrow beam by their passage through lead bricks. The beam was allowed to fall on a thin foil of gold of thickness 2.1 × 10–7 m. The scattered alpha-particles were observed through a rotatable detector consisting of zinc sulphide screen and a microscope. The scattered alpha-particles on striking the screen produced brief light flashes or scintillations. These flashes may be FIGURE 12.1 Geiger-Marsden scattering experiment. viewed through a microscope and the The entire apparatus is placed in a vacuum chamber distribution of the number of scattered (not shown in this figure). particles may be studied as a function of angle of scattering. 416 FIGURE 12.2 Schematic arrangement of the Geiger-Marsden experiment. A typical graph of the total number of α-particles scattered at different angles, in a given interval of time, is shown in Fig. 12.3. The dots in this figure represent the data points and the solid curve is the theoretical prediction based on the assumption that the target atom has a small, dense, positively charged nucleus. Many of the α-particles pass through the foil. It means that they do not suffer any collisions. Only about 0.14% of the incident α-particles scatter by more than 1°; and about 1 in 8000 deflect by more than 90°. Rutherford argued that, to deflect the α-particle backwards, it must experience a large repulsive force. This force could 2020-21

Atoms be provided if the greater part of the mass of the atom and its positive charge were concentrated tightly at its centre. Then the incoming α-particle could get very close to the positive charge without penetrating it, and such a close encounter would result in a large deflection. This agreement supported the hypothesis of the nuclear atom. This is why Rutherford is credited with the discovery of the nucleus. In Rutherford’s nuclear model of the atom, the entire positive charge and most of the mass of the atom are concentrated in the nucleus with the electrons some distance away. The electrons would be moving in orbits FIGURE 12.3 Experimental data points (shown by dots) on scattering of α-particles by a thin foil at about the nucleus just as the planets different angles obtained by Geiger and Marsden do around the sun. Rutherford’s using the setup shown in Figs. 12.1 and experiments suggested the size of 12.2. Rutherford’s nuclear model predicts the solid the nucleus to be about 10–15 m to 10–14 m. From kinetic theory, the size curve which is seen to be in good agreement with of an atom was known to be 10–10 m, experiment. about 10,000 to 100,000 times larger than the size of the nucleus (see Chapter 11, Section 11.6 in Class XI Physics textbook). Thus, the electrons would seem to be at a distance from the nucleus of about 10,000 to 100,000 times the size of the nucleus itself. Thus, most of an atom is empty space. With the atom being largely empty space, it is easy to see why most α-particles go right through a thin metal foil. However, when α-particle happens to come near a nucleus, the intense electric field there scatters it through a large angle. The atomic electrons, being so light, do not appreciably affect the α-particles. The scattering data shown in Fig. 12.3 can be analysed by employing Rutherford’s nuclear model of the atom. As the gold foil is very thin, it can be assumed that α-particles will suffer not more than one scattering during their passage through it. Therefore, computation of the trajectory of an alpha-particle scattered by a single nucleus is enough. Alpha- particles are nuclei of helium atoms and, therefore, carry two units, 2e, of positive charge and have the mass of the helium atom. The charge of the gold nucleus is Ze, where Z is the atomic number of the atom; for gold Z = 79. Since the nucleus of gold is about 50 times heavier than an α-particle, it is reasonable to assume that it remains stationary throughout the scattering process. Under these assumptions, the trajectory of an alpha-particle can be computed employing Newton’s second law of motion and the Coulomb’s law for electrostatic force of repulsion between the alpha-particle and the positively 417 charged nucleus. 2020-21

Physics The magnitude of this force is 1 (2e )( Ze ) (12.1) F = 4πε0 r2 where r is the distance between the α-particle and the nucleus. The force is directed along the line joining the α-particle and the nucleus. The magnitude and direction of the force on an α-particle continuously changes as it approaches the nucleus and recedes away from it. 12.2.1 Alpha-particle trajectory The trajectory traced by an α-particle depends on the impact parameter, b of collision. The impact parameter is the perpendicular distance of the initial velocity vector of the α-particle from the centre of the nucleus (Fig. 12.4). A given beam of α-particles has a distribution of impact parameters b, so that the beam is scattered in various directions with different probabilities (Fig. 12.4). (In a beam, all particles have nearly same kinetic energy.) It is seen that an α-particle close to the nucleus (small impact parameter) suffers large scattering. In case of head-on collision, the impact parameter is minimum and the α-particle rebounds back (θ ≅ π). For a large impact parameter, the α-particle goes nearly undeviated and FIGURE 12.4 Trajectory of α-particles in the has a small deflection (θ ≅ 0). coulomb field of a target nucleus. The impact The fact that only a small fraction of the parameter, b and scattering angle θ number of incident particles rebound back are also depicted. indicates that the number of α-particles undergoing head on collision is small. This, in turn, implies that the mass and positive charge of the atom is concentrated in a small volume. Rutherford scattering therefore, is a powerful way to determine an upper limit to the size of the nucleus. 418 EXAMPLE 12.1 Example 12.1 In the Rutherford’s nuclear model of the atom, the nucleus (radius about 10–15 m) is analogous to the sun about which the electron move in orbit (radius ≈ 10–10 m) like the earth orbits around the sun. If the dimensions of the solar system had the same proportions as those of the atom, would the earth be closer to or farther away from the sun than actually it is ? The radius of earth’s orbit is about 1.5 × 1011 m. The radius of sun is taken as 7 × 108 m. Solution The ratio of the radius of electron’s orbit to the radius of nucleus is (10–10 m)/(10–15 m) = 105, that is, the radius of the electron’s orbit is 105 times larger than the radius of nucleus. If the radius of the earth’s orbit around the sun were 105 times larger than the radius of the sun, the radius of the earth’s orbit would be 105 × 7 × 108 m = 7 × 1013 m. This is more than 100 times greater than the actual orbital radius of earth. Thus, the earth would be much farther away from the sun. It implies that an atom contains a much greater fraction of empty space than our solar system does. 2020-21

Atoms Example 12.2 In a Geiger-Marsden experiment, what is the distance of closest approach to the nucleus of a 7.7 MeV α-particle before it comes momentarily to rest and reverses its direction? Solution The key idea here is that throughout the scattering process, the total mechanical energy of the system consisting of an α-particle and a gold nucleus is conserved. The system’s initial mechanical oαteTonf-heeiptrthasgireyntmiiitscseilyaceEslh.ti,aTeennhmbieeec.fraofgTilryneheanEetlheiepreingsoepytrajeguErnytsftitciEwalfethlhieaseennnjdkutehisrnntegeuyttαchil-cUeepuaeescrlnetaeiicnncrtlgrteeiybcrmeaKpcooctom,tafelaenctnnthuidtaelaalrititienleynicdseosremftgqroyiuopnmasUg.l Eq. (12.1). Let d be the centre-to-centre distance between the α-particle and Simulate Rutherford scattering experiment the gold nucleus when the α-particle is at its stopping point. Then http://www-outreach.phy.cam.ac.uk/camphy/nucleus/nucleus6_1.htm we can write the conservation of energy Ei = Ef as EXAMPLE 12.2 K = 1 (2e )(Ze) = 2Ze 2 4πε0 d 4πε0d Thus the distance of closest approach d is given by d = 2 Ze 2 4πε0K The maximum kinetic energy found in α-particles of natural origin is 7.7 MeV or 1.2 × 10–12 109 m2/C2. with e = 1.6 × 10–19 C, Jw.eShinacvee,1/4πε0 = 9.0 × N Therefore d = (2)(9.0 × 109 Nm2 /C 2 )(1.6 × 10–19 C )2 Z 1.2 × 10−12 J = 3.84 × 10–16 Z m The atomic number of foil material gold is Z = 79, so that d (Au) = 3.0 × 10–14 m = 30 fm. (1 fm (i.e. fermi) = 10–15 m.) The radius of gold nucleus is, therefore, less than 3.0 × 10–14 m. This is not in very good agreement with the observed result as the actual radius of gold nucleus is 6 fm. The cause of discrepancy is that the distance of closest approach is considerably larger than the sum of the radii of the gold nucleus and the α-particle. Thus, the α-particle reverses its motion without ever actually touching the gold nucleus. 12.2.2 Electron orbits The Rutherford nuclear model of the atom which involves classical concepts, pictures the atom as an electrically neutral sphere consisting of a very small, massive and positively charged nucleus at the centre surrounded by the revolving electrons in their respective dynamically sketleeaecbptlreotohnresbmaitnsi.dnTththheeeenilreuccotlrrebouiststsap.triTcohvfoiudrsce,esfotohfraeatrteraqdcuytiinsoiantme, Fciceeanbltelrytiwpseetetaanbltlfehoerocrreebvi(otFlcvi)nintago hydrogen atom Fe = Fc 1 e2 = mv2 (12.2) 419 4πε0 r2 r 2020-21

Physics Thus the relation between the orbit radius and the electron velocity is r = e2 (12.3) 4 πε 0mv 2 The kinetic energy (K ) and electrostatic potential energy (U ) of the electron in hydrogen atom are K = 1 mv 2 = e2 and U = − e2 2 8πε0r 4πε0r (The negative sign in U signifies that the electrostatic force is in the –r direction.) Thus the total energy E of the electron in a hydrogen atom is E = K +U = e2 − e2 8πε0r 4πε0r = − e2 (12.4) 8πε0r The total energy of the electron is negative. This implies the fact that the electron is bound to the nucleus. If E were positive, an electron will not follow a closed orbit around the nucleus. Example 12.3 It is found experimentally that 13.6 eV energy is required to separate a hydrogen atom into a proton and an electron. Compute the orbital radius and the velocity of the electron in a hydrogen atom. Solution Total energy of the electron in hydrogen atom is –13.6 eV = –13.6 × 1.6 × 10–19 J = –2.2 ×10–18 J. Thus from Eq. (12.4), we have E = − e2 = −2.2 × 10−18 J 8πε0r This gives the orbital radius r = − e2 = − (9 × 109 N m2/C2 )(1.6 × 10−19C)2 8πε0 (2)(–2.2 × 10−18 J) EXAMPLE 12.3 E = 5.3 × 10–11 m. The velocity of the revolving electron can be computed from Eq. (12.3) with m = 9.1 × 10–31 kg, v= e = 2.2 × 106 m/s. 4πε0mr 420 12.3 ATOMIC SPECTRA As mentioned in Section 12.1, each element has a characteristic spectrum of radiation, which it emits. When an atomic gas or vapour is excited at low pressure, usually by passing an electric current through it, the emitted radiation has a spectrum which contains certain specific wavelengths only. A spectrum of this kind is termed as emission line spectrum and it 2020-21

Atoms consists of bright lines on a dark background. The spectrum emitted by atomic hydrogen is shown in Fig. 12.5. Study of emission line spectra of a material can therefore serve as a type of “fingerprint” for identification of the gas. When white light passes through a gas and we analyse the transmitted light using a spectrometer we find some dark lines in the FIGURE 12.5 Emission lines in the spectrum of hydrogen. spectrum. These dark lines correspond precisely to those wavelengths which were found in the emission line spectrum of the gas. This is called the absorption spectrum of the material of the gas. 12.3.1 Spectral series We might expect that the frequencies of the light emitted by a particular element would exhibit some regular pattern. Hydrogen is the simplest atom and therefore, has the simplest spectrum. In the observed spectrum, however, at first sight, there does not seem to be any resemblance of order or regularity in spectral lines. But the spacing between lines within certain sets of the hydrogen spectrum decreases in a regular way (Fig. 12.5). Each of these sets is called a spectral series. In 1885, the first such series was observed by a Swedish school teacher Johann Jakob Balmer (1825–1898) in the visible region of the hydrogen spectrum. This series is called Balmer series (Fig. 12.6). The line with the longest wavelength, 656.3 nm in the red is called Hα; the FIGURE 12.6 Balmer series in the next line with wavelength 486.1 nm in the blue- emission spectrum of hydrogen. green is called HHβγ,; the third line 434.1 nm in the violet is called and so on. As the wavelength decreases, the lines appear closer together and are weaker in intensity. Balmer found a simple empirical formula for the observed wavelengths 1 1 1 (12.5) λ = R  22 − n 2  where λ is the wavelength, R is a constant called the Rydberg constant, and n may have integral values 3, 4, 5, etc. The value of R is 1.097 × 107 m–1. This equation is also called Balmer formula. Taking n = 3 in Eq. (12.5), one obtains the wavelength of the Hα line: 1 = 1.097 × 107 1 1 m –1 λ  22 − 32  = 1.522 × 106 m–1 421 i.e., λ = 656.3 nm 2020-21

Physics For n = 4, one obtains tahteλw=a3ve6l4e.n6gnthmo.fTHhβislinise,thetecs. hFoorrtnes=t ∞, one obtains the limit of the series, wavelength in the Balmer series. Beyond this limit, no further distinct lines appear, instead only a faint continuous spectrum is seen. Other series of spectra for hydrogen were subsequently discovered. These are known, after their discoverers, as Lyman, Paschen, Brackett, and Pfund series. These are represented by the formulae: Lyman series: 1 =R 1 − 1  n = 2,3,4... (12.6) λ  12 n2  Paschen series: 1 = R  1 − 1  n = 4,5,6... (12.7) λ  32 n2  Brackett series: 1 =R  1 − 1  n = 5,6,7... (12.8) λ  42 n2  Pfund series: 1 =R 1 1  n = 6,7,8... (12.9) λ  52 − n2  The Lyman series is in the ultraviolet, and the Paschen, Brackett, and Pfund series are in the infrared region. The Balmer formula Eq. (12.5) may be written in terms of frequency of the light, recalling that c = νλ or 1=ν λc Thus, Eq. (12.5) becomes ν = Rc 1 − 1 (12.10)  22 n 2  There are only a few elements (hydrogen, singly ionised helium, and doubly ionised lithium) whose spectra can be represented by simple formula like Eqs. (12.5) – (12.9). Equations (12.5) – (12.9) are useful as they give the wavelengths that hydrogen atoms radiate or absorb. However, these results are empirical and do not give any reasoning why only certain frequencies are observed in the hydrogen spectrum. 12.4 BOHR MODEL OF THE HYDROGEN ATOM The model of the atom proposed by Rutherford assumes that the atom, consisting of a central nucleus and revolving electron is stable much like sun-planet system which the model imitates. However, there are some fundamental differences between the two situations. While the planetary 422 system is held by gravitational force, the nucleus-electron system being charged objects, interact by Coulomb’s Law of force. We know that an 2020-21

object which moves in a circle is being constantly Atoms NIELS HENRIK DAVID BOHR (1885 – 1962) accelerated – the acceleration being centripetal in nature. According to classical electromagnetic theory, an Niels Henrik David Bohr (1885 – 1962) Danish accelerating charged particle emits radiation in the form physicist who explained the of electromagnetic waves. The energy of an accelerating spectrum of hydrogen atom electron should therefore, continuously decrease. The based on quantum ideas. He gave a theory of nuclear electron would spiral inward and eventually fall into the fission based on the liquid- nucleus (Fig. 12.7). Thus, such an atom can not be stable. drop model of nucleus. Further, according to the classical electromagnetic theory, Bohr contributed to the clarification of conceptual the frequency of the electromagnetic waves emitted by problems in quantum the revolving electrons is equal to the frequency of mechanics, in particular by revolution. As the electrons spiral inwards, their angular proposing the comple- mentary principle. velocities and hence their frequencies would change continuously, and so will the frequency of the light emitted. Thus, they would emit a continuous spectrum, in contradiction to the line spectrum actually observed. Clearly Rutherford model tells only a part of the story implying that the classical ideas are not sufficient to explain the atomic structure. FIGURE 12.7 An accelerated atomic electron must spiral into the nucleus as it loses energy. Example 12.4 According to the classical electromagnetic theory, calculate the initial frequency of the light emitted by the electron revolving around a proton in hydrogen atom. Solution From Example 12.3 we know that velocity of electron moving around a proton in hydrogen atom in an orbit of radius 5.3 × 10–11 m is 2.2 × 10–6 m/s. Thus, the frequency of the electron moving around the proton is v 2.2 × 106 m s−1 2πr 2π 5.3 × 10−11 m =( )ν = ≈ 6.6 × 1015 Hz. EXAMPLE 12.4 423 According to the classical electromagnetic theory we know that the frequency of the electromagnetic waves emitted by the revolving electrons is equal to the frequency of its revolution around the nucleus. Thus the initial frequency of the light emitted is 6.6 × 1015 Hz. 2020-21

Physics It was Niels Bohr (1885 – 1962) who made certain modifications in this model by adding the ideas of the newly developing quantum hypothesis. Niels Bohr studied in Rutherford’s laboratory for several months in 1912 and he was convinced about the validity of Rutherford nuclear model. Faced with the dilemma as discussed above, Bohr, in 1913, concluded that in spite of the success of electromagnetic theory in explaining large-scale phenomena, it could not be applied to the processes at the atomic scale. It became clear that a fairly radical departure from the established principles of classical mechanics and electromagnetism would be needed to understand the structure of atoms and the relation of atomic structure to atomic spectra. Bohr combined classical and early quantum concepts and gave his theory in the form of three postulates. These are : (i) Bohr’s first postulate was that an electron in an atom could revolve in certain stable orbits without the emission of radiant energy, contrary to the predictions of electromagnetic theory. According to this postulate, each atom has certain definite stable states in which it can exist, and each possible state has definite total energy. These are called the stationary states of the atom. (ii) Bohr’s second postulate defines these stable orbits. This postulate states that the electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of h/2π where h is the Planck’s constant (= 6.6 × 10–34 J s). Thus the angular momentum (L) of the orbiting electron is quantised. That is L = nh/2π (12.11) (iii) Bohr’s third postulate incorporated into atomic theory the early quantum concepts that had been developed by Planck and Einstein. It states that an electron might make a transition from one of its specified non-radiating orbits to another of lower energy. When it does so, a photon is emitted having energy equal to the energy difference between the initial and final states. The frequency of the emitted photon is then given by hν = Ei – Ef (12.12) wheFroerEai ahnyddErof gaerentahteomen,eErgqi.es(1o2f.t4h)egiinveitsiatlhaenedxfpinreaslssitoantetsoadnedteErim>iEnfe. the energies of different energy states. But then this equation requires the radius r of the electron orbit. To calculate r, Bohr’s second postulate about the angular momentum of the electron–the quantisation condition – is used. The angular momentum L is given by L = mvr Bohr’s second postulate of quantisation [Eq. (12.11)] says that the allowed values of angular momentum are integral multiples of h/2π. Ln = mvnrn = nh (12.13) 2π 424 where n misoavninignetleegcetrr,onrniins tthheenrtah doirubsit.oTf hnethapllooswseidbloerobritbsitaraennduvmn ibserthede speed of 2020-21

Atoms 1, 2, 3 ..., according to the values of n, which is called the principal quantum number of the orbit. From Eq. (12.3), the relation between vn and rn is vn = e 4πε0mrn Combining it with Eq. (12.13), we get the following expressions for vn and rn, vn = 1 e2 1 (12.14) n 4πε0 (h 2π) and rn = n2   h 2 4πε0 (12.15)  m   2π e2 Eq. (12.14) depicts that the orbital speed in the nth orbit falls by a factor of n. Using Eq. (12.15), the size of the innermost orbit (n = 1) can be obtained as r1 = h 2ε0 πme 2 This is called the Bohr radius, represented by the symbol a0. Thus, a0 = h 2ε0 (12.16) πme 2 Substitution of values of h, m, ε0 and e gives a0 = 5.29 × 10–11 m. From Eq. (12.15), it can also be seen that the radii of the orbits increase as n2. The total energy of the electron in the stationary states of the hydrogen atom can be obtained by substituting the value of orbital radius in Eq. (12.4) as  e2   m   2hπ  2  e2   8πε0   n2    4πε0  En = − or En = − me 4 2 (12.17) 8n 2ε02h Substituting values, Eq. (12.17) yields En = − 2.18 × 10−18 J (12.18) n2 Atomic energies are often expressed in electron volts (eV) rather than joules. Since 1 eV = 1.6 × 10–19 J, Eq. (12.18) can be rewritten as 13.6 eV (12.19) En = − n 2 The negative sign of the total energy of an electron moving in an orbit means that the electron is bound with the nucleus. Energy will thus be required to remove the electron from the hydrogen atom to a distance 425 infinitely far away from its nucleus (or proton in hydrogen atom). 2020-21

Physics The derivation of Eqs. (12.17) – (12.19) involves the assumption that the electronic orbits are circular, though orbits under inverse square force are, in general elliptical. (Planets move in elliptical orbits under the inverse square gravitational force of the sun.) However, it was shown by the German physicist Arnold Sommerfeld (1868 – 1951) that, when the restriction of circular orbit is relaxed, these equations continue to hold even for elliptic orbits. ORBIT VS STATE (ORBITAL PICTURE) OF ELECTRON IN ATOM We are introduced to the Bohr Model of atom one time or the other in the course of physics. This model has its place in the history of quantum mechanics and particularly in explaining the structure of an atom. It has become a milestone since Bohr introduced the revolutionary idea of definite energy orbits for the electrons, contrary to the classical picture requiring an accelerating particle to radiate. Bohr also introduced the idea of quantisation of angular momentum of electrons moving in definite orbits. Thus it was a semi-classical picture of the structure of atom. Now with the development of quantum mechanics, we have a better understanding of the structure of atom. Solutions of the Schrödinger wave equation assign a wave-like description to the electrons bound in an atom due to attractive forces of the protons. An orbit of the electron in the Bohr model is the circular path of motion of an electron around the nucleus. But according to quantum mechanics, we cannot associate a definite path with the motion of the electrons in an atom. We can only talk about the probability of finding an electron in a certain region of space around the nucleus. This probability can be inferred from the one-electron wave function called the orbital. This function depends only on the coordinates of the electron. It is therefore essential that we understand the subtle differences that exist in the two models: Bohr model is valid for only one-electron atoms/ions; an energy value, assigned to each orbit, depends on the principal quantum number n in this model. We know that energy associated with a stationary state of an electron depends on n only, for one-electron atoms/ions. For a multi-electron atom/ion, this is not true. The solution of the Schrödinger wave equation, obtained for hydrogen-like atoms/ ions, called the wave function, gives information about the probability of finding an electron in various regions around the nucleus. This orbital has no resemblance whatsoever with the orbit defined for an electron in the Bohr model. 426 EXAMPLE 12.5 Example 12.5 A 10 kg satellite circles earth once every 2 h in an orbit having a radius of 8000 km. Assuming that Bohr’s angular momentum postulate applies to satellites just as it does to an electron in the hydrogen atom, find the quantum number of the orbit of the satellite. Solution From Eq. (12.13), we have m vn rn = nh/2π 2020-21

Atoms Here m = 10 kg aansd2rhn .= 8 × 106 m. We have the time period T of the circling satellite That is T = 7200 s. TThheusquthaentvuemlocnituymvnbe=r 2π rtnh/eT.orbit of satellite nSu=b(s2tπitrunt)2in×g of m/(T × h). EXAMPLE 12.5 the values, n = (2π × 8 × 106 m)2 × 10/(7200 s × 6.64 × 10–34 J s) = 5.3 × 1045 Note that the quantum number for the satellite motion is extremely large! In fact for such large quantum numbers the results of quantisation conditions tend to those of classical physics. 12.4.1 Energy levels The energy of an atom is the least (largest negative value) when its electron is revolving in an orbit closest to the nucleus i.e., the one for which n = 1. For n = 2, 3, ... the absolute value of the energy E is smaller, hence the energy is progressively larger in the outer orbits. The lowest state of the atom, called the ground state, is that of the lowest energy, with the electron revolving in the orbit of smallest radius, the Bohr radius, a 0. The energy of this state (n = 1), E1 is –13.6 eV. Therefore, the minimum energy required to free the electron from the ground state of the hydrogen atom is 13.6 eV. It is called the ionisation energy of the hydrogen atom. This prediction of the Bohr’s model is in excellent agreement with the experimental value of ionisation energy. At room temperature, most of the hydrogen atoms are in ground state. When a hydrogen atom receives energy by processes such as electron collisions, the atom may acquire sufficient energy to raise the electron to higher energy states. The atom is then said to be in an excited state. From Eq. (12.19), for n = r2e;quthireedentoeregxycitEe2 is –3.40 eV. It means that the energy an electron in hydrogen atom to its first excited state, is an FIGURE 12.8 The energy level diagram for the hydrogen atom. etShnimeerhiglayydrleryqo,ugEea3nl=tao–t1oE.m251–freEoV1m=aint–sd3.gE4r30o–ueEnV1d–=s(1t–a21t.3e0.(96n)e=eVV1, )o=tro1tos0ee.2cxoceniVtde. The electron in a hydrogen atom excited state (n = 3), 12.09 eV energy is required, and so at room temperature spends most of its time in the ground on. From these excited states the electron can then fall back state. To ionise a hydrogen to a state of lower energy, emitting a photon in the process. atom an electron from the Thus, as the excitation of hydrogen atom increases (that is ground state, 13.6 eV of energy must be supplied. (The horizontal as n increases) the value of minimum energy required to lines specify the presence of free the electron from the excited atom decreases. allowed energy states.) The energy level diagram* for the stationary states of a hydrogen atom, computed from Eq. (12.19), is given in * An electron can have any total energy above E = 0 eV. In such situations the 427 electron is free. Thus there is a continuum of energy states above E = 0 eV, as shown in Fig. 12.8. 2020-21

Physics Fig. 12.8. The principal quantum number n labels the stationary states in the ascending order of energy. In this diagram, the highest energy state corresponds to n = ∞ in Eq, (12.19) and has an energy of 0 eV. This is the energy of the atom when the electron is completely removed (r = ∞) from the nucleus and is at rest. Observe how the energies of the excited states come closer and closer together as n increases. FRANCK – HERTZ EXPERIMENT The existence of discrete energy levels in an atom was directly verified in 1914 by James Franck and Gustav Hertz. They studied the spectrum of mercury vapour when electrons having different kinetic energies passed through the vapour. The electron energy was varied by subjecting the electrons to electric fields of varying strength. The electrons collide with the mercury atoms and can transfer energy to the mercury atoms. This can only happen when the energy of the electron is higher than the energy difference between an energy level of Hg occupied by an electron and a higher unoccupied level (see Figure). For instance, the difference between an occupied energy level of Hg and a higher unoccupied level is 4.9 eV. If an electron of having an energy of 4.9 eV or more passes through mercury, an electron in mercury atom can absorb energy from the bombarding electron and get excited to the higher level [Fig (a)]. The colliding electron’s kinetic energy would reduce by this amount. The excited electron would subsequently fall back to the ground state by emission of radiation [Fig. (b)]. The wavelength of emitted radiation is: λ = hc = 6.625 ×10−34 × 3 ×108 = 253 nm E 4.9 ×1.6 ×10−19 By direct measurement, Franck and Hertz found that the emission spectrum of mercury has a line corresponding to this wavelength. For this experimental verification of Bohr’s basic ideas of discrete energy levels in atoms and the process of photon emission, Frank and Hertz were awarded the Nobel prize in 1925. 12.5 THE LINE SPECTRA OF THE HYDROGEN ATOM According to the third postulate of Bohr’s model, when an atom makes a transition from the higher energy state with quantum number ni to the leonweergryenisecrgayrrsietdataewwaiythbyquaapnhtuotmonnoufmfrbeeqruennf c(nyf < nsui ),chthtehdatifference of 428 νif 2020-21

Atoms hνif = Eni – Enf (12.20) Using Eq. (12.16), for Enf and Eni, we get hνif = me 4  1 − 1 (12.21)   8ε02h 2 n 2 n 2 f i or νif = me 4  1 1 (12.22)  − ni2  8ε02h 3 n 2 f Equation (12.21) is the Rydberg formula, for the spectrum of the hydrogen atom. In this rteolEatqi.o(n1,2i.f1w0)efotraktheenBf a=lm2earnsderineis=. T3h,e4R,y5d.b..e,rigt reduces to a form similar constant R is readily identified to be me 4 (12.23) R = 8ε02h 3c If we insert the values of various constants in Eq. (12.23), we get R = 1.03 × 107 m–1 This is a value very close to the value (1.097 × 107 m–1) obtained from the empirical Balmer formula. This agreement between the theoretical and experimental values of the Rydberg constant provided a direct and striking confirmation of the Bohr’s model. t h Since both tnef layndshnoi are integers, is immedia ws that in transitions between different atomic levels, light is radiated in various discrete frequencies. For hydrogen spectrum, the Balmer formula ecotcr.reTshpeonredssutlotsnof =f t2haenBdonhir=’s3m, 4o,d5e,l suggested the presence of other series spectra for hydrogen atom–those corresponding to transitions resulting safrenordmiensniwf==er41e, a5n, detnc.i ,=a2n,d3s, oetocn.;. nSfu=ch3 identified in the course of spectroscopic investigations and are known as the Lyman, Balmer, Paschen, Brackett, and Pfund series. The electronic transitions corresponding to these series are shown in Fig. 12.9. The various lines in the atomic spectra are produced when electrons jump from higher energy state to a lower energy state and photons are emitted. These spectral lines are called emission lines. But when an atom FIGURE 12.9 Line spectra originate in 429 absorbs a photon that has precisely transitions between energy levels. 2020-21

Physics the same energy needed by the electron in a lower energy state to make transitions to a higher energy state, the process is called absorption. Thus if photons with a continuous range of frequencies pass through a rarefied gas and then are analysed with a spectrometer, a series of dark spectral absorption lines appear in the continuous spectrum. The dark lines indicate the frequencies that have been absorbed by the atoms of the gas. The explanation of the hydrogen atom spectrum provided by Bohr’s model was a brilliant achievement, which greatly stimulated progress towards the modern quantum theory. In 1922, Bohr was awarded Nobel Prize in Physics. Example 12.6 Using the Rydberg formula, calculate the wavelengths of the first four spectral lines in the Lyman series of the hydrogen spectrum. Solution The Rydberg formula is hc/λif = me 4  1 1  − ni2  8ε02h 2 n 2 f The wavelengths of the first four lines in the Lyman series correspond to transitions from ni = 2,3,4,5 to nf = 1. We know that me 4 = 13.6 eV = 21.76 ×10–19 J 8ε02h 2 Therefore, λi1 = hc 1 1 m × 10−19  1 21.76 − n 2  i EXAMPLE 12.6 = 6.625 ×10−34 × 3 ×108 × ni2 m = 0.9134 ni2 ×10−7 m 21.76 ×10−19 × (ni2 −1) (ni2 −1) Su=bs9ti1tu3t.i4ngnin2i/(=ni22 ,–31,4) ,Å5, we get λ21 = 1218 Å, λ31 = 1028 Å, λ41 = 974.3 Å, and λ51 = 951.4 Å. 12.6 DE BROGLIE’S EXPLANATION OF BOHR’S SECOND POSTULATE OF QUANTISATION Of all the postulates, Bohr made in his model of the atom, perhaps the most puzzling is his second postulate. It states that the angular momentum of the electron orbiting around the nucleus is quantised (that oisn,lLynth=onshe/v2aπlu; ens = 1, 2, 3 …). Why should the angular momentum have that are integral multiples of h/2π? The French physicist Louis de Broglie explained this puzzle in 1923, ten years after Bohr proposed his model. We studied, in Chapter 11, about the de Broglie’s hypothesis that 430 material particles, such as electrons, also have a wave nature. C. J. Davisson and L. H. Germer later experimentally verified the wave nature of electrons 2020-21

Atoms in 1927. Louis de Broglie argued that the electron in its circular orbit, as proposed by Bohr, must be seen as a particle wave. In analogy to waves travelling on a string, particle waves too can lead to standing waves under resonant conditions. From Chapter 15 of Class XI Physics textbook, we know that when a string is plucked, a vast number of wavelengths are excited. However only those wavelengths survive which have nodes at the ends and form the standing wave in the string. It means that in a string, standing waves are formed when the total distance travelled by a wave down the string and back is one wavelength, two wavelengths, or any integral number of wavelengths. Waves with other wavelengths interfere with themselves upon reflection and their amplitudes quickly drop to zero. For an electron moving in nth circular orbit of radius rTnh, uthse total distance is the circumference of the orbit, 2πrn. FIGURE 12.10 A standing wave is shown on a circular 2π rn = nλ, n = 1, 2, 3... (12.24) orbit where four de Broglie Figure 12.10 illustrates a standing particle wave on a wavelengths fit into the circular orbit for n = 4, i.e., m2πorvni=ng4λin, wnhtheorrebλiti.sFtrhoemdeCBhraopgtleier circumference of the orbit. wavelength of the electron 11, we have λ = h/p, where p is the magnitude of the electron’s momentum. If the speed of the electron is much less than the speed of light, the momentum is mvn. Thus, λ = h/mvn. From Eq. (12.24), we have 2π rn = n h/mvn or m vn rn = nh/2π This is the quantum condition proposed by Bohr for the angular momentum of the electron [Eq. (12.13)]. In Section 12.5, we saw that this equation is the basis of explaining the discrete orbits and energy levels in hydrogen atom. Thus de Broglie hypothesis provided an explanation for Bohr’s second postulate for the quantisation of angular momentum of the orbiting electron. The quantised electron orbits and energy states are due to the wave nature of the electron and only resonant standing waves can persist. Bohr’s model, involving classical trajectory picture (planet-like electron orbiting the nucleus), correctly predicts the gross features of the hydrogenic atoms*, in particular, the frequencies of the radiation emitted or selectively absorbed. This model however has many limitations. Some are: (i) The Bohr model is applicable to hydrogenic atoms. It cannot be extended even to mere two electron atoms such as helium. The analysis of atoms with more than one electron was attempted on the lines of Bohr’s model for hydrogenic atoms but did not meet with any success. Difficulty lies in the fact that each electron interacts not only with the positively charged nucleus but also with all other electrons. * Hydrogenic atoms are the atoms consisting of a nucleus with positive charge 431 +Ze and a single electron, where Z is the proton number. Examples are hydrogen atom, singly ionised helium, doubly ionised lithium, and so forth. In these atoms more complex electron-electron interactions are nonexistent. 2020-21

Physics The formulation of Bohr model involves electrical force between positively charged nucleus and electron. It does not include the electrical forces between electrons which necessarily appear in multi-electron atoms. (ii) While the Bohr’s model correctly predicts the frequencies of the light emitted by hydrogenic atoms, the model is unable to explain the relative intensities of the frequencies in the spectrum. In emission spectrum of hydrogen, some of the visible frequencies have weak intensity, others strong. Why? Experimental observations depict that some transitions are more favoured than others. Bohr’s model is unable to account for the intensity variations. Bohr’s model presents an elegant picture of an atom and cannot be generalised to complex atoms. For complex atoms we have to use a new and radical theory based on Quantum Mechanics, which provides a more complete picture of the atomic structure. LASER LIGHT Imagine a crowded market place or a railway platform with people entering a gate and going towards all directions. Their footsteps are random and there is no phase correlation between them. On the other hand, think of a large number of soldiers in a regulated march. Their footsteps are very well correlated. See figure here. This is similar to the difference between light emitted by an ordinary source like a candle or a bulb and that emitted by a laser. The acronym LASER stands for Light Amplification by Stimulated Emission of Radiation. Since its development in 1960, it has entered into all areas of science and technology. It has found applications in physics, chemistry, biology, medicine, surgery, engineering, etc. There are low power lasers, with a power of 0.5 mW, called pencil lasers, which serve as pointers. There are also lasers of different power, suitable for delicate surgery of eye or glands in the stomach. Finally, there are lasers which can cut or weld steel. Light is emitted from a source in the form of packets of waves. Light coming out from an ordinary source contains a mixture of many wavelengths. There is also no phase relation between the various waves. Therefore, such light, even if it is passed through an aperture, spreads very fast and the beam size increases rapidly with distance. In the case of laser light, the wavelength of each packet is almost the same. Also the average length of the packet of waves is much larger. This means that there is better phase correlation over a longer duration of time. This results in reducing the divergence of a laser beam substantially. If there are N atoms in a source, each emitting light with intensity I, then the total intensity produced by an ordinary source is proportional to NI, whereas in a laser source, it is proportional to N2I. Considering that N is very large, we see that the light from a laser can be much stronger than that from an ordinary source. When astronauts of the Apollo missions visited the moon, they placed a mirror on its surface, facing the earth. Then scientists on the earth sent a strong laser beam, which was reflected by the mirror on the moon and received back on the earth. The size of the reflected laser beam and the time taken for the round trip were measured. This allowed a very accurate determination of (a) the extremely small divergence of a laser beam and (b) the distance of the moon from the earth. 432 2020-21

Atoms SUMMARY 1. Atom, as a whole, is electrically neutral and therefore contains equal amount of positive and negative charges. 2. In Thomson’s model, an atom is a spherical cloud of positive charges with electrons embedded in it. 3. In Rutherford’s model, most of the mass of the atom and all its positive charge are concentrated in a tiny nucleus (typically one by ten thousand the size of an atom), and the electrons revolve around it. 4. Rutherford nuclear model has two main difficulties in explaining the structure of atom: (a) It predicts that atoms are unstable because the accelerated electrons revolving around the nucleus must spiral into the nucleus. This contradicts the stability of matter. (b) It cannot explain the characteristic line spectra of atoms of different elements. 5. Atoms of most of the elements are stable and emit characteristic spectrum. The spectrum consists of a set of isolated parallel lines termed as line spectrum. It provides useful information about the atomic structure. 6. The atomic hydrogen emits a line spectrum consisting of various series. The frequency of any line in a series can be expressed as a difference of two terms; Lyman series: ν= Rc  1 − 1  ; n = 2, 3, 4,...  12 n2  Balmer series: ν= Rc  1 − 1  ;n = 3, 4, 5,...  22 n2  Paschen series: ν= Rc  1 − 1  ; n = 4, 5, 6,...  32 n2  Brackett series: ν= Rc  1 − 1  ; n = 5, 6, 7,...  42 n2  Pfund series: ν= Rc  1 − 1  ; n = 6, 7, 8,...  52 n2  7. To explain the line spectra emitted by atoms, as well as the stability of atoms, Niel’s Bohr proposed a model for hydrogenic (single elctron) atoms. He introduced three postulates and laid the foundations of quantum mechanics: (a) In a hydrogen atom, an electron revolves in certain stable orbits (called stationary orbits) without the emission of radiant energy. (b) The stationary orbits are those for which the angular momentum is some integral multiple of h/2π. (Bohr’s quantisation condition.) That is L = nh/2π, where n is an integer called the principal quantum number. (c) The third postulate states that an electron might make a transition from one of its specified non-radiating orbits to another of lower energy. When it does so, a photon is emitted having energy equal to the energy difference between the initial and final states. The frequency (ν) of the emitted photon is then given by hν = Ei – Ef An atom absorbs radiation of the same frequency the atom emits, in which case the electron is transferred to an orbit with a higher value of n. Ei + hν = Ef 433 2020-21

Physics 8. As a result of the quantisation condition of angular momentum, the electron orbits the nucleus at only specific radii. For a hydrogen atom it is given by rn = n2  h 2 4πε0  m   2π e2 The total energy is also quantised: En = − me 4 2 8n 2ε02h = –13.6 eV/n2 The n = 1 state is called ground state. In hydrogen atom the ground state energy is –13.6 eV. Higher values of n correspond to excited states (n > 1). Atoms are excited to these higher states by collisions with other atoms or electrons or by absorption of a photon of right frequency. 9. de Broglie’s hypothesis that electrons have a wavelength λ = h/mv gave an explanation for Bohr’s quantised orbits by bringing in the wave- particle duality. The orbits correspond to circular standing waves in which the circumference of the orbit equals a whole number of wavelengths. 10. Bohr’s model is applicable only to hydrogenic (single electron) atoms. It cannot be extended to even two electron atoms such as helium. This model is also unable to explain for the relative intensities of the frequencies emitted even by hydrogenic atoms. 434 POINTS TO PONDER 1. Both the Thomson’s as well as the Rutherford’s models constitute an unstable system. Thomson’s model is unstable electrostatically, while Rutherford’s model is unstable because of electromagnetic radiation of orbiting electrons. 2. What made Bohr quantise angular momentum (second postulate) and not some other quantity? Note, h has dimensions of angular momentum, and for circular orbits, angular momentum is a very relevant quantity. The second postulate is then so natural! 3. The orbital picture in Bohr’s model of the hydrogen atom was inconsistent with the uncertainty principle. It was replaced by modern quantum mechanics in which Bohr’s orbits are regions where the electron may be found with large probability. 4. Unlike the situation in the solar system, where planet-planet gravitational forces are very small as compared to the gravitational force of the sun on each planet (because the mass of the sun is so much greater than the mass of any of the planets), the electron-electron electric force interaction is comparable in magnitude to the electron- nucleus electrical force, because the charges and distances are of the same order of magnitude. This is the reason why the Bohr’s model with its planet-like electron is not applicable to many electron atoms. 5. Bohr laid the foundation of the quantum theory by postulating specific orbits in which electrons do not radiate. Bohr’s model include only 2020-21

Atoms one quantum number n. The new theory called quantum mechanics supportes Bohr’s postulate. However in quantum mechanics (more generally accepted), a given energy level may not correspond to just one quantum state. For example, a state is characterised by four quantum numbers (n, l, m, and s), but for a pure Coulomb potential (as in hydrogen atom) the energy depends only on n. 6. In Bohr model, contrary to ordinary classical expectation, the frequency of revolution of an electron in its orbit is not connected to the frequency of spectral line. The later is the difference between two orbital energies divided by h. For transitions between large quantum numbers (n to n – 1, n very large), however, the two coincide as expected. 7. Bohr’s semiclassical model based on some aspects of classical physics and some aspects of modern physics also does not provide a true picture of the simplest hydrogenic atoms. The true picture is quantum mechanical affair which differs from Bohr model in a number of fundamental ways. But then if the Bohr model is not strictly correct, why do we bother about it? The reasons which make Bohr’s model still useful are: (i) The model is based on just three postulates but accounts for almost all the general features of the hydrogen spectrum. (ii) The model incorporates many of the concepts we have learnt in classical physics. (iii) The model demonstrates how a theoretical physicist occasionally must quite literally ignore certain problems of approach in hopes of being able to make some predictions. If the predictions of the theory or model agree with experiment, a theoretician then must somehow hope to explain away or rationalise the problems that were ignored along the way. EXERCISES 12.1 Choose the correct alternative from the clues given at the end of 12.2 the each statement: (a) The size of the atom in Thomson’s model is .......... the atomic 435 size in Rutherford’s model. (much greater than/no different from/much less than.) (b) In the ground state of .......... electrons are in stable equilibrium, while in .......... electrons always experience a net force. (Thomson’s model/ Rutherford’s model.) (c) A classical atom based on .......... is doomed to collapse. (Thomson’s model/ Rutherford’s model.) (d) An atom has a nearly continuous mass distribution in a .......... but has a highly non-uniform mass distribution in .......... (Thomson’s model/ Rutherford’s model.) (e) The positively charged part of the atom possesses most of the mass in .......... (Rutherford’s model/both the models.) Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect? 2020-21

Physics 12.3 What is the shortest wavelength present in the Paschen series of spectral lines? 12.4 A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level? 12.5 The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state? 12.6 A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon. 12.7 (a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels. (b) Calculate the orbital period in each of these levels. 12.8 The radius of the innermost electron orbit of a hydrogen atom is 5.3×10–11 m. What are the radii of the n = 2 and n =3 orbits? 12.9 A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted? 12.10 In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius 1.5 × 1011 m with orbital speed 3 × 104 m/s. (Mass of earth = 6.0 × 1024 kg.) 436 ADDITIONAL EXERCISES 12.11 Answer the following questions, which help you understand the difference between Thomson’s model and Rutherford’s model better. (a) Is the average angle of deflection of α-particles by a thin gold foil predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model? (b) Is the probability of backward scattering (i.e., scattering of α-particles at angles greater than 90°) predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model? (c) Keeping other factors fixed, it is found experimentally that for small thickness t, the number of α-particles scattered at moderate angles is proportional to t. What clue does this linear dependence on t provide? (d) In which model is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering of α-particles by a thin foil? 12.12 The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about 10–40. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting. 12.13 Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n–1). For large n, show that this frequency equals the classical frequency of revolution of the electron in the orbit. 2020-21

Atoms 12.14 Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size? Why is an atom not, say, thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom (~ 10–10m). (a) Construct a quantity with the dimensions of length from the fundamental constants e, me, and c. Determine its numerical value. (b) You will find that the length obtained in (a) is many orders of magnitude smaller than the atomic dimensions. Further, it involves c. But energies of atoms are mostly in non-relativistic domain where c is not expected to play any role. This is what may have suggested Bohr to discard c and look for ‘something else’ to get the right atomic size. Now, the Planck’s constant h had already made its appearance elsewhere. Bohr’s great insight lay in recognising that h, mwiet,hanthde e will yield the right atomic size. Construct a quantity dimension of length from h, me, and e and confirm that its numerical value has indeed the correct order of magnitude. 12.15 The total energy of an electron in the first excited state of the hydrogen atom is about –3.4 eV. (a) What is the kinetic energy of the electron in this state? (b) What is the potential energy of the electron in this state? (c) Which of the answers above would change if the choice of the zero of potential energy is changed? 12.16 If Bohr’s quantisation postulate (angular momentum = nh/2π) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why then do we never speak of quantisation of orbits of planets around the sun? 12.17 Obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom [i.e., an atom in which a negatively charged muon (µ–) of mass about 207me orbits around a proton]. 437 2020-21

Physics Chapter Thirteen NUCLEI 13.1 INTRODUCTION In the previous chapter, we have learnt that in every atom, the positive charge and mass are densely concentrated at the centre of the atom forming its nucleus. The overall dimensions of a nucleus are much smaller than those of an atom. Experiments on scattering of α-particles demonstrated that the radius of a nucleus was smaller than the radius of an atom by a factor of about 104. This means the volume of a nucleus is about 10–12 times the volume of the atom. In other words, an atom is almost empty. If an atom is enlarged to the size of a classroom, the nucleus would be of the size of pinhead. Nevertheless, the nucleus contains most (more than 99.9%) of the mass of an atom. Does the nucleus have a structure, just as the atom does? If so, what are the constituents of the nucleus? How are these held together? In this chapter, we shall look for answers to such questions. We shall discuss various properties of nuclei such as their size, mass and stability, and also associated nuclear phenomena such as radioactivity, fission and fusion. 13.2 ATOMIC MASSES AND COMPOSITION OF NUCLEUS The mass of an atom is very small, compared to a kilogram; for example, the mass of a carbon atom, 12C, is 1.992647 × 10–26 kg. Kilogram is not 438 a very convenient unit to measure such small quantities. Therefore, a 2020-21

Nuclei different mass unit is used for expressing atomic masses. This unit is the atomic mass unit (u), defined as 1/12th of the mass of the carbon (12C) atom. According to this definition 1u = mass of one 12C atom 12 = 1.992647 ×10−26 kg 12 = 1.660539 ×10−27 kg (13.1) The atomic masses of various elements expressed in atomic mass unit (u) are close to being integral multiples of the mass of a hydrogen atom. There are, however, many striking exceptions to this rule. For example, the atomic mass of chlorine atom is 35.46 u. Accurate measurement of atomic masses is carried out with a mass spectrometer, The measurement of atomic masses reveals the existence of different types of atoms of the same element, which exhibit the same chemical properties, but differ in mass. Such atomic species of the same element differing in mass are called isotopes. (In Greek, isotope means the same place, i.e. they occur in the same place in the periodic table of elements.) It was found that practically every element consists of a mixture of several isotopes. The relative abundance of different isotopes differs from element to element. Chlorine, for example, has two isotopes having masses 34.98 u and 36.98 u, which are nearly integral multiples of the mass of a hydrogen atom. The relative abundances of these isotopes are 75.4 and 24.6 per cent, respectively. Thus, the average mass of a chlorine atom is obtained by the weighted average of the masses of the two isotopes, which works out to be = 75.4 × 34.98 + 24.6 × 36.98 100 = 35.47 u which agrees with the atomic mass of chlorine. Even the lightest element, hydrogen has three isotopes having masses 1.0078 u, 2.0141 u, and 3.0160 u. The nucleus of the lightest atom of hydrogen, which has a relative abundance of 99.985%, is called the proton. The mass of a proton is m p = 1.00727 u = 1.67262 × 10−27 kg (13.2) This is equal to the mass of the hydrogen atom (= 1.00783u), minus the mass of a single electron (me = 0.00055 u). The other two isotopes of hydrogen are called deuterium and tritium. Tritium nuclei, being unstable, do not occur naturally and are produced artificially in laboratories. The positive charge in the nucleus is that of the protons. A proton carries one unit of fundamental charge and is stable. It was earlier thought that the nucleus may contain electrons, but this was ruled out later using arguments based on quantum theory. All the electrons of an atom are outside the nucleus. We know that the number of these electrons outside 439 the nucleus of the atom is Z, the atomic number. The total charge of the 2020-21