Xam Idea Physics - Class 12 Term 1 and 2 Question Bank

9. The figure shows a modified Young’s double slit experimental set-up. Here SS2 – SS1 = λ/4. 2 S1 P S O S2 (a) Write the condition for constructive interference. (b) Obtain an expression for the fringe width. 10. (a) If one of two identical slits producing interference in Young’s experiment is covered with glass, so that the light intensity passing through it is reduced to 50%, find the ratio of the maximum and minimum intensity of the fringe in the interference pattern. (b) What kind of fringes do you expect to observe if white light is used instead of monochromatic light? 3 11. Answer the following: (a) In what way is diffraction from each slit related to the interference pattern in a double slit experiment? (b) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain, why. (c) How does the resolving power of a microscope depend on (i) the wavelength of the light used and (ii) the medium used between the object and the objective lens? 3 12. (a) Using the phenomenon of polarisation, show how transverse nature of light can be demonstrated. (b) Two polaroids P1 and P2 are placed with their pass axes perpendicular to each other. Unpolarised light of intensity I0 is incident on P1. A third polaroid P3 is kept in between P1 and P2 such that its pass axis makes an angle of 30° with that of P1. Determine the intensity of light transmitted through P1, P2 and P3. 3 13. Describe diffraction of light due to a single slit. Explain formation of a pattern of fringes obtained on the screen and plot showing variation of intensity with angle θ in single slit diffraction. 5 Answers 1. (i) (b) (ii) (b) (iii) (d) 2. (i) plane (ii) limit of resolution 8. 2.4 ×10–4m 12. I0 , 3I0 , 3I0 2 32 8 zzz 448 Xam idea Physics–XII

Dual Nature Chapter –11 of Matter and Radiation 1. Dual Nature of Radiations It is well known that the phenomena of interference, diffraction and polarisation indicate that light has wave nature. But some phenomena like photoelectric effect, Compton effect, emission and absorption of radiation could not be explained by wave nature. These were explained by particle (quantum) nature of light. Thus, light (radiation) has dual nature. 2. Quantum Nature of Light: Concept of a Photon Some phenomena like photoelectric effect, Compton effect, Raman effect could not be explained by wave theory of light. Therefore, quantum theory of light was proposed by Einstein. According to quantum theory of light “light is propagated in bundles of small energy, each bundle being called a photon and possessing energy.” E = hν = hc ...(i) m where ν is frequency, λ is wavelength of light and h is Planck’s constant = 6.62 × 10–34 joule second and c = speed of light in vacuum = 3 × 108 m/s. Momentum of photon, p= hν = h ...(ii) c m Rest mass of photon = 0 Dynamic or kinetic mass of photon, m= hν = h ...(iii) c2 cm 3. Photoelectric Effect The phenomenon of emission of electrons from a metallic surface by the use of light (or radiant) energy is called photoelectric effect. The phenomenon was discovered by Lenard. For photoelectric emission, the metal used must have low work function, e.g., alkali metals. Caesium is the best metal for photoelectric effect. 4. Hertz’s Observations The phenomenon of photoelectric effect was discovered by Heinrich Hertz in 1887. While performing an experiment for production of electromagnetic waves by means of spark discharge, Hertz observed that sparks occurred more rapidly in the air gap of his transmitter when ultraviolet radiations was directed at one of the metal plates. Hertz could not explain his observations. 5. Lenard’s Observations Ultraviolet mA Phillip Lenard observed that when ultraviolet radiations radiations were made incident on the emitter plate of an evacuated CA glass tube enclosing two metal plates (called electrodes), current flows in the circuit, but as soon as ultraviolet radiation falling on the emitter plate was stopped, the current flow stopped. These observations indicate that when ultraviolet radiations fall on the emitter (cathode) –+ Dual Nature of Matter and Radiation 449

plate C, the electrons are ejected from it, which are attracted towards anode plate A. The electrons flow through the evacuated glass tube, complete the circuit and current begins to flow in the circuit. Hallwachs Exp.: Hallwachs studied further by taking a zinc Ultraviolet Zinc plate plate and an electroscope. The zinc plate was connected to an rays electroscope. He observed that: (i) When an uncharged zinc plate was irradiated by ultraviolet Electron light, the zinc plate acquired positive charge. (ii) When a positively charged zinc plate is illuminated by e– ultraviolet light, the positive charge of the plate was increased. Gold leaf (iii) When a negatively charged zinc plate was irradiated by electroscope ultraviolet light, the zinc plate lost its charge. All these observations show that when ultraviolet light falls on zinc plate, the negatively charged particles (electrons) are emitted. Further study done by Hallwach’s experiment shows that different metals emit electrons by different electromagnetic radiations. For example the alkali metals (e.g., sodium, caesium, potassium etc.) emit electrons when visible light is incident on them. The heavy metals (such as zinc, cadmium, magnesium etc.) emit electrons when ultraviolet radiation is incident on them. Caesium is the most sensitive metal for photoelectric emission. It can emit electrons with less- energetic infrared radiation. In photoelectric effect the light energy is converted into electrical energy. 6. Characteristics of Photoelectric Effect (i) Effect of Intensity: Intensity of light means the energy incident per unit area per second. For a given frequency, if intensity of incident light is increased, the photoelectric current increases and with decrease of intensity, the photoelectric current decreases; but the stopping potential remains the same. Intensity of radiations can be increased/decreased by varying the distance between source and metal plate (or emitter). Current (I) ν3 > ν2 > ν1 Current (I) 3I 2I ν3 ν2 ν1 I Vs3 Vs2 Vs1 O – (V) + (V) –Vs O (b) Potential difference (V) (a) This means that the intensity of incident light affects the photoelectric current but the maximum kinetic energy of photoelectrons remains unchanged as shown in fig (b). (ii) Effect of Frequency: When the intensity of incident light is kept fixed and frequency is increased, the photoelectric current remains the same; but the stopping potential increases. If the frequency is decreased, the stopping potential decreases and at a particular frequency of incident light, the stopping potential becomes zero. This value of frequency of incident light for which the stopping potential is zero is called threshold frequency ν0. If the frequency of incident light (ν) is less than the threshold frequency (ν0) no photoelectric emission takes place. Thus, the increase of frequency increases the maximum kinetic energy of photoelectrons but the photoelectric current remain unchanged. 450 Xam idea Physics–XII

(iii) Effect of Photometal: When frequency and intensity of Metal Metal incident light are kept fixed and photometal is changed, 1 2 we observe that stopping potential (VS) versus frequency Vs ( 0)1 ( 0)2 Frequency (ν) graphs are parallel straight lines, cutting frequency O axis at different points (Fig.). This shows that threshold frequencies are different for different metals, the slope (VS / ν) for all the metals is same and hence a universal constant. (iv) Effect of Time: There is no time lag between the incidence of light and the emission of photoelectrons. 7. Some Definitions Work Function: The minimum energy required to free an electron from its metallic bonding is called work function. It is denoted by W or φ and is usually expressed in electron volt (l eV = 1.6 × 10–19 J). Threshold Frequency: The minimum frequency of incident light which is just capable of ejecting electrons from a metal is called the threshold frequency. It is denoted by It is different for different metal. ν0. Stopping Potential: The minimum retarding potential applied to anode of a photoelectric tube which is just capable of stopping photoelectric current is called the stopping potential. It is denoted by V0 (or VS) 8. Einstein’s Explanation of Photoelectric Effect: Einstein’s Photoelectric Equation Einstein extended Planck’s quantum idea for light to explain photoelectric effect. The assumptions of Einstein’s theory are: Photometal Light G 1. The photoelectric effect is the result of collision of a photon of incident light and an electron of photometal. –+ 2. The electron of photometal is bound with the nucleus by Photoelectric cell R coulomb attractive forces. The minimum energy required to free an electron from its bondage is called work function (W). 3. The incident photon interacts with a single electron and loses its energy in two parts: (i) in releasing the electron from its bondage, and (ii) in imparting kinetic energy to emitted electron. Accordingly, if hν is the energy of incident photon, then from law of conservation of energy kinetic energy of hν = W + Ek 1 mvm2 ax = hν – W ...(i) or maximum photoelectrons, Ek = 2 where W is work function. This equation is referred as Einstein’s photoelectric equation and explains all experimental results of photoelectric effect. If Vs is stopping potential, then Ek = 1 mvm2 ax = eVS …(ii) 2 Stopping potential, Vs = h ν – W …(iii) e e The slope of Ek versus ν graph is h. h The slope of VS versus ν graph is e . 9. Photocell A photocell is a device which converts light energy into electrical energy. It is also called electric eye. 10. Matter Waves: Wave Nature of Particles Light exhibits particle aspects in certain phenomena (e.g., photoelectric effect, emission and absorption of radiation), while wave aspects in other phenomena (e.g., interference, diffraction Dual Nature of Matter and Radiation 451

and polarisation). That is, light has dual nature. In analogy with dual nature of light, de Broglie thought in terms of dual nature of matter. 11. de Broglie Hypothesis Louis de Broglie postulated that the material particles (e.g., electrons, protons, α-particles, atoms, etc.) may exhibit wave aspect. Accordingly, a moving material particle behaves as wave and the wavelength associated with material particle is m= h = h , where p is momentum. p mv If Ek is kinetic energy of moving material particle, then p = 2mEk m= h 2mEk h h h i.e., m= p = mv = 2mEk The wave associated with material particle is called the de-Broglie wave or matter wave. The de-Broglie hypothesis has been confirmed by diffraction experiments. For charged particles associated through a potential of V volt, Ek = qV m= h 2mqV For electrons, q = e =1.6 × 10–19 C, m = 9 ×10–31 kg m = 12.27 # 10–10 m = 12.27 Å (Only for electrons) VV h h For electron orbiting in an atom, de Broglie wavelength is given as m= p = mv For neutral particles in thermal equilibrium at absolute temperature T, Ek = kT m= h 2mkT 12. Davisson and Germer Experiment This experiment gave the first experimental evidence for the wave nature of slow electrons. Later on, it was shown that all material particles in motion behave as waves. Selected NCERT Textbook Questions Photoelectric Effect Q. 1. Find the (a) maximum frequency and (b) minimum wavelength of X-rays produced by 30 kV electrons. Ans. Given V = 30 kV = 30 × 103 volt Energy, E = eV = 1.6×10–19 × 30 × 103= 4.8 ×10–15 joule (a) Maximum frequency nmax is given by, E = hnmax νmax = E = 4.8 ×10 –15 = 7.24 ×1018 Hz h 6.63 ×10 –34 3×108 (b) Minimum wavelength, mmin = c = 7.24×1018 = 4.1×10–11 m = 0.041 nm νmax 452 Xam idea Physics–XII

Q. 2. The work function of caesium metal is 2.14 eV. When light of frequency 6×1014 Hz is incident on the metal surface, photoemission of electrons occurs. What is the (a) maximum kinetic energy of the emitted electrons? (b) stopping potential and (c) maximum speed of emitted electrons? Ans. Given φ0 = 2.14 eV, ν = 6×10–14 Hz (a) Maximum kinetic energy of emitted electron Ek = hν – φ0 = 6.63 × 10–34 × 6 × 1014 – 2.14 × 1.6 × 10–19 = 0.554 # 10–19 J= 0.554 # 10–19 eV = 0.34 eV 1.6 # 10–19 (b) Stopping potential V0 is given by Ek = eV0 & V0 = Ek = 0.34 eV = 0.34 V e e (c) Maximum speed (vmax) of emitted electrons is given by 1 mvm2 ax = Ek 2 or vmax = 2Ek = 2 # 0.554 # 10–19 = 3.48×105 m/s m 9.1 # 10–31 Q. 3. The photoelectric cut-off voltage in a certain photoelectric experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted? Ans. Cut-off voltage, V0 = 1.5 V Maximum kinetic energy of photoelectrons Ek = eV0 = 1.5 eV = 1.5 × 1.6 × 10–19 J = 2.4× 10–19 J Q. 4. The energy flux of sunlight reaching the surface of 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. Ans. Energy of each photon E= hc = 6.63 # 10–34 # 3 # 108 = 3.62 # 10–19 J m 550 # 10–9 Number of photons incident on earth’s surface per second per square metre = Total energy per square metre per second Energy of one photon = 1.388 # 103 = 3.8 # 1021 3.62 # 10–19 Q. 5. In an experiment of photoelectric effect, the slope of cut-off voltage versus frequency of incident light is found to be 4.12×10–15 Vs. Calculate the value of Planck’s constant. Ans. Einstein’s photoelectric equation is Ek = hν – φ0 or eV0 = hν – φ0 or V0 = h ν – z0 e e h Clearly slope of V0 – ν curve is e Give h = 4.12×10–15 Vs & h = 4.12×10–15 eVs e = 4.12×10–15 × 1.6 × 10–19 = 6.59×10–34 Js Dual Nature of Matter and Radiation 453

Q. 6. A 100 W 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 associated per photon with sodium light? (b) At what rate are the photons delivered to the sphere? Ans. Given P = 100 W, λ = 589 nm = 589× 10– 9 m (a) Energy of one photon E= hc = 6.63 # 10–34 # 3 # 108 = 3.38 # 10–19 J m 589 # 10–9 (b) Number of photons (n) delivered to the sphere per second is given by P = nE & n = P = 100 = 3.0 # 1020 photons/ second E 3.38 # 10–19 Q. 7. 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 photoelectric emission. Ans. Einstein’s photoelectric equation is hν = hν0 + Ek or hν = hν0 + eV0 eV0 = h(ν – ν0) = 6.63×10–34 (8.2×1014 –3.3×1014) joule or cut-off voltage V0 = 6.63 # 4.9 # 10–20 = 6.63 # 4.9 # 10–20 V = 2.03 V e 1.6 # 10–19 Q. 8. The work function of a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm? Ans. The energy of incident radiations E = hc = 6.63 # 10–34 # 3 # 108 joule = 6.03 ×10 –19 joule m 330 # 10–9 = 6.03 # 10–19 eV = 3.77 eV 1.6 # 10–19 The work function of photometal, φ0 = 4.2 eV As energy of incident photon is less than work function, photoemission is not possible. Q. 9. Light of frequency 7.21×1014 Hz is incident on a metal surface. Electrons with a maximum speed of 6.0×105 ms–1 are ejected from the surface. What is the threshold frequency for photoemission of electrons? (Planck’s constant h = 6.62×10– 34 Js) Ans. Given ν = 7.21×1014 Hz, vmax = 6.0×105 ms–1 From Einstein’s photoelectric equation Ek = hν – hν0, where ν0 is the threshold frequency ν0 = ho – Ek = dν – Ek n = ν – 1 mvm2 ax h h 2 h = 7.21 # 1014 – 9.1 # 10–31 # (6.0 # 105)2 2 # 6.62 # 10–34 = 7.21 × 1014 – 2.47 × 1014 = 4.74×1014 Hz Q . 10. 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 cathode the stopping potential of photoelectrons is 0.38 V. Find the work function of the cathode material. 454 Xam idea Physics–XII

Ans. Given λ = 488 nm = 488 × 10–9 m, V0 = 0.38 V, φ0 =? hc Energy of incident photon E= m        = 6.63 # 10–34 # 3 # 108 = 4.08×10–19 J 488 # 10–9 4.08 # 10–19        = 1.6 # 10–19 = 2.55 eV From Einstein’s photoelectric equation hc = z0 + eV0 m hc Work function z0 = m – eV0 = 2.55 eV – 0.38 eV = 2.17 eV Q. 11. The work function of 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 a 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? Ans. Energy of incident photon, E = hc m Here λ = 3300 Å = 3300×10–10 m = 3.3×10–7 m E = 6.63 # 10–34 # 3 # 108 joule 3.3 # 10–7 = 6.63 # 10–34 # 3 # 108 eV = 3.76 eV 3.3 # 10–7 # 1.6 # 10–19 Photoelectric emission is only possible if energy of incident photon is equal to or greater than the work function. For Na and K this condition is satisfied, hence photoelectric emission is possible; but in the case of Mo and Ni, the energy of incident photon is less than the work function; hence photoelectric emission is not possible. If source is brought nearer, then the intensity of incident radiation increases but frequency of a photon remains the same; therefore Mo and Ni will still not show photoelectric effect; however in the case of Na and K the current will increase in same proportion as the increase in intensity takes place. de Broglie Waves Q. 12. Calculate the (a) momentum and (b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V. Ans. For electron, mass m = 9.1×10–31 kg (a) Momentum p = 2mEk = 2meV = 2 # 9.1 # 10–31 # 1.6 # 10–19 # 56 = 4.04 ×10–24 kg ms–1 (b) de Broglie wavelength m = h = 6.63 # 10–34 = 1.64 # 10–10 m = 0.164 nm p 4.04 # 10–24 Q. 13. What is the (a) momentum (b) speed and (c) de Broglie wavelength of an electron with kinetic energy of 120 eV? (mass of electron me = 9.1×10–31 kg, h = 6.63×10–34 Js) Ans. Given kinetic energy, Ek = 120 eV = 120×1.6×10–19 J = 1.92×10–17 J (a) Momentum of electron, p = 2me Ek        = 2 # 9.1 # 10–31 # 1.92 # 10–17 = 5.91×10–24 kg ms–1 Dual Nature of Matter and Radiation 455

(b) Speed of electron, v= p = 5.91 # 10–24 = 6.5×106 m/s m 9.1 # 10–31 (c) de Broglie wavelength, m= h = 6.63 # 10–34 = 1.12 # 10–10 m = 0.112 nm p 5.91 # 10–24 Q. 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. [CBSE Guwahati 2015] Ans. Given λ = 589 nm = 5.89×10–7 m The de Broglie wavelength m= h = h & m2 = h2 p 2mEk 2mEk ∴ Kinetic energy Ek = h2 2mm2 (a) For electron Ek = h2 2mm2 (6.63 # 10–34)2 ∴ Ek = 2 # 9.1 # 10–31 # (5.89 # 10–7)2 = 6.96×10–25 J (b) For neutron m = 1.67×10–27 kg Ek = (6.63 # 10–34)2 = 3.79 ×10–28 J 2 # 1.67 # 10–27 # (5.89 # 10–7)2 Q. 15. What is the de Broglie wavelength of: (a) a bullet of mass 0.040 kg travelling at a speed of 1.0 km/s. (b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s. (c) a dust particle of mass 1.0×10–9 kg drifting with a speed of 2.2 m/s. Ans. (a) m= h = h = 6.63×10–34 = 1.66×10–35 m p mv 0.040×1.0×103 (b) m= h = 6.63×10–34 = 1.1×10–32 m mv 0.060×1.0 (c) m = h = 6.63×10–34 = 3.01×10–25 m mv 1.0×10–9 ×2.2 Obviously de Broglie wavelength decreases with increase of momentum. Q. 16. An electron and a photon, each has a wavelength of 1.00 nm. Find (a) their momenta (b) the energy of the photon and (c) the kinetic energy of electron. [CBSE Delhi 2011] Ans. Given λ = 1.00 nm = 1.00×10–9 m (a) Momenta of electron and photon are equal; given by p = h = 6.63 # 10–34 = 6.63×10–25 kg ms–1 m 1.00 # 10–9 (b) Energy of photon, E = hν = h. c = h c m m     = pc = 6.63 # 10–25 # 3 # 108 J = 19.89 ×10–17 J = 19.89 # 10–17 eV =1.24 ×103 eV = 1.24 keV 1.6 # 10–19 456 Xam idea Physics–XII

(c) Kinetic energy of electron Ek = 1 me v2 = p2 2 2me = (6.63 # 10–25)2 J 2 # 9.1 # 10–31 = 2.42 # 10–19 J = 2.42 # 10–19 eV = 1.51 eV 1.6 # 10–19 Q. 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, 3 having an average kinetic energy 2 kT at 300 K. [Mass of neutron=1.67×10–27 kg] Ans. (a) de Broglie’s wavelength m = h 2mEk Kinetic energy Ek = h2 = (6.63×10 –34) 2 = 6.7×10–21 J 2mm2 2×1.67×10–27 × (1.40×10–10)2 (b) m = h= h = h 2mEk 3mkT 2m× 3 kT 2 = 6.63×10–34 3×1.67×10–27 ×1.38×10–23 ×300    = 6.63×10–34 = 1.46×10–10 m = 0.146 nm 4.55×10–24 Q. 18. Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon). Ans. Momentum of a photon of frequency ν (wavelength λ) is given by p= ho = h c m ∴ Wavelength of electromagnetic radiation m= h p ∴ de Broglie wavelength m= h p Thus wavelength of electromagnetic radiation is equal to de Broglie wavelength of its quantum. Q . 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. Ans. Root mean square speed, vrms = 3kT m Mass of nitrogen molecule, m = 2×14.0076 = 28.0152 u = 28.0152×1.66×10–27 kg de Broglie wavelength, m = h = h = h mvrms 3mkT m. 3kT m 6.63 # 10–34 = 3 # 28.0152 # 1.66 # 10–27 # 1.38 # 10–23 # 300 = 6.63 # 10–34 = 2.76 # 10–11 m = 0.276 Ac 2.40 # 10–23 Dual Nature of Matter and Radiation 457

Q. 20. 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 (λy = 5.9×10–7 m)? [CBSE (AI) 2014] Ans. de Broglie wavelength associated with electron m = 12.27 # 10–10 m V Here V = 50 kV = 50×103 V ∴ m= 12.27 ×10–10 = 5.5 # 10–12 m 50 # 103 Wavelength of yellow light, λy = 5.9×10–7 m The resolving power of an electron microscope is given by RP = 1 = 2n sin b dmin 1.22m Where dmin = minimum separation For constant numerical aperture 1 m Resolving power of microscope ? ∴ Resolving power of electron microscope = my = 5.9 # 10–7 . 105 Resolving power of optical microscope m 5.5 # 10–12 That is, resolving power of electron microscope is 105 times the resolving power of optical microscope. Multiple Choice Questions [1 mark] Choose and write the correct option(s) in the following questions. 1. A particle is dropped from a height H. The de Broglie wavelength of the particle as a function of height is proportional to [NCERT Exemplar] (b) H1/2 (a) H (c) H0 (d) H –1/2 2. The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with 1 MeV energy is nearly [NCERT Exemplar] (b) 1.2 × 10–3 nm (a) 1.2 nm (c) 1.2 × 10–6 nm (d) 1.2 × 101 nm 3. Consider a beam of electrons (each electron with energy E0) incident on a metal surface kept in an evacuated chamber. Then [NCERT Exemplar] (a) no electrons will be emitted as only photons can emit electrons. (b) electrons can be emitted but all with an energy, E0. (c) electrons can be emitted with any energy, with a maximum of E0 – f (f is the work function). (d) electrons can be emitted with any energy, with a maximum of E0. 4. The threshold wavelength for photoelectric emission from a material is 5200 Å. Photoelectrons will be emitted when this material is illuminated with monochromatic radiation from a: (a) 50 watt infrared lamp (b) 1000 watt infrared lamp (c) 1 watt ultraviolet lamp (d) 1 watt infrared lamp 458 Xam idea Physics–XII

5. A photoelectric cell is illuminated by a point source of light 1 m away. The plate emits electrons having stopping potential V. Then: (a) V decreases as distance increase (b) V increases as distance increase (c) V is independent of distance (r) (d) V becomes zero when distance increases or decreases 6. In a photoelectric experiment, the stopping- potential for the incident light of wavelength 4000 Å is 2 volt. If the wavelength be changed to 3000 Å, the stopping-potential will be: (a) 2 volt (b) less than 2 volt (c) zero (d) more than 2 volt. 7. The work-function for a metal is 3 eV. To emit a photoelectron of energy 2 eV from the surface of this metal, the wavelength of the incident light should be: (a) 6187 Å (b) 4125 Å (c) 12375 Å (d) 2486 Å 8. In the Davisson and Germer experiment, the velocity of electrons emitted from the electron gun can be increased by (a) increasing the potential difference between the anode and filament (b) increasing the filament current (c) decreasing the filament current (d) decreasing the potential difference between the anode and filament 9. The work-function of a surface of a photosensitive material is 6.2 eV. The wavelength of incident radiation for which the stopping potential is 5 V lies in: (a) ultraviolet region (b) visible region (c) infrared region (d) X-ray region 10. A proton, a neutron, an electron and an a-particle have same energy. Then their de Broglie wavelengths compare as [NCERT Exemplar] (a) m p = mn > me > ma (b) ma < m p = mn > me (c) me < m p = mn > ma (d) me = m p = mn = ma 11. The number of photoelectrons emitted for light of frequency ν (higher than the threshold frequency ν0) is proportional to: (a) threshold frequency (b) intensity of light (c) frequency of light (d) ν – ν0 12. Relativistic corrections become necessary when the expression for the kinetic energy 1 2 mv2 , becomes comparable with mc2, where m is the mass of the particle. At what de Broglie wavelength will relativistic corrections become important for an electron? [NCERT Exemplar] (a) λ = 10 nm (b) λ = 10–1 nm (c) λ = 10–4 nm (d) λ = 10–6 nm 13. Monochromatic light of wavelength 667 nm is produced by a helium neon laser. The power emitted is 9 mW. The number of photons arriving per second on the average at a target irradiated by this beam is: (a) 3 × 1016 (b) 9 × 1015 (c) 3 × 1019 (d) 9 × 1017 14. Electrons used in an electron microscope are accelerated by a voltage of 25 kV. If the voltage is increased to 100 kV then the de-Broglie wavelength associated with the electrons would (a) increase by 2 times (b) decrease by 2 times (c) decrease by 4 times (d) increase by 4 times Dual Nature of Matter and Radiation 459

15. Two particles A1 and A2 of masses m1, m2 (m1 > m2) have the same de Broglie wavelength. Then [NCERT Exemplar] (a) their momenta are the same (b) their energies are the same (c) energy of A1 is less than the energy of A2 (d) energy of A1 is more than the energy of A2 16. An electron (mass m) with an initial velocity v = v0it is in an electric field E = E0 tj. If m0 = h/mv0 , it’s de Breoglie wavelength at time t is given by [NCERT Exemplar] (a) m0 (b) m0 1+ e2 E02 t2 (c) m0 (d) m0 m2 v02 1+ e2 E02 t2 f1 + e2 E02 t2 p m2 v02 m2 v02 17. If an electron and a photon propagate in the form of waves having same wavelength, it implies that they have same: (a) speed (b) momentum (c) energy (d) all the above 18. A particle of mass 1 mg has the same wavelength as an electron moving with a velocity of 3 × 106 ms –1. The velocity of the particle is (mass of electron = 9.1 × 10–31 kg) (a) 2.7 × 10–18 ms–1 (b) 9 × 10–2 ms–1 (c) 3 × 10–31 ms–1 (d) 2.7 × 10–21 ms–1 19. Which of the following figures represent the variation of particle momentum and the associated de-Broglie wavelength? (a) (b) (c) (d) Kinetic energy p p Kinetic energy p Kinetic energy p Kinetic energy λλλλ 20. According to Einstein's photoelectric equation, the graph between the kinetic energy of photoelectrons ejected and the frequency of incident radiation is (a) (b) (c) (d) Frequency Frequency Frequency Frequency Answers 2. (b) 3. (d) 4. (c) 5. (c) 6. (d) 8. (a) 9. (a) 10. (b) 11. (b) 12. (c), (d) 1. (d) 14. (b) 15. (a), (c) 16. (c) 17. (b) 18. (a) 7. (d) 20. (d) 13. (a) 19. (d) 460 Xam idea Physics–XII

Fill in the Blanks [1 mark] 1. The minimum energy required by a free electron to just escape from the metal surface is called as _________________. 2. The maximum kinetic energy of emitted photoelectrons depends on the _________________ of incident radiation and the nature of material. 3. The velocity of photon in different media is _________________. 4. The main aim of Davisson-Germer experiment is to verify the _________________ nature of moving electrons. 5. The minimum frequency required to eject an electron from the surface of a metal surface is called _________________ frequency. 6. In photoelectric effect, saturation current is not affected on decreasing the _________________ of incident radiation provided its intensity remains unchanged. 7. The intensity of radiation also depends upon the number of _________________. 8. Momentum of photon in different media is _________________. 9. Davisson and Germer experiment established the _________________. 10. Matter wave are associated with _________________ particle. Answers 1. work function 2. frequency 3. different 4. wave 5. threshold 6. wavelength/frequency 7. photons 8. different 9. wave nature 10. moving Very Short Answer Questions [1 mark] Q. 1. Name the phenomenon which shows the quantum nature of electromagnetic radiation. [CBSE (AI) 2017] Ans. “Photoelectric effect” shows the quantum nature of electromagnetic radiation. Q. 2. Define intensity of radiation on the basis of photon picture of light. Write its SI unit. [CBSE (AI) 2014; 2019 (55/1/1)] Ans. The amount of light energy or photon energy incident per metre square per second is called intensity of radiation. SI unit: W or J/s – m2 PQ m2 Q. 3. The figure shows the variation of stopping potential V0 with the frequency n of the incident radiations for two photosensitive metals P and Q. Which metal has smaller V0 threshold wavelength? Justify your answer. ν  [CBSE 2019 (55/4/1)] 0.1 1.0 (× 1015 s–1) Ans. Since m0 = c , metal Q has smaller threshold wavelength. o0 Q. 4. Write the basic features of photon picture of electromagnetic radiation on which Einstein’s photoelectric equation is based. [CBSE Delhi 2013] Ans. Features of the photons: p = h , where h is (i) Photons are particles of light having energy E = hν and momentum m Planck constant. (ii) Photons travel with the speed of light in vacuum, independent of the frame of reference. (iii) Intensity of light depends on the number of photons crossing unit area in a unit time. Dual Nature of Matter and Radiation 461

Q. 5. Define the term ‘stopping potential’ in relation to photoelectric effect. [CBSE (AI) 2011] Ans. The minimum retarding (negative) potential of anode of a photoelectric tube for which photoelectric current stops or becomes zero is called the stopping potential. Q. 6. Define the term ‘threshold frequency’ in relations to photoelectric effects. [CBSE (F) 2011, 2019 (55/1/1)] Ans. Threshold frequency is defined as the minimum frequency of incident radiation which can cause photoelectric emission. It is different for different metal. Q. 7. In photoelectric effect, why should the photoelectric current increase as the intensity of monochromatic radiation incident on a photosensitive surface is increased? Explain. [CBSE (F) 2014] Ans. The photoelectric current increases proportionally with the increase in intensity of incident radiation. Larger the intensity of incident radiation, larger is the number of incident photons and hence larger is the number of electrons ejected from the photosensitive surface. Q. 8. Write the expression for the de Broglie wavelength associated with a charged particle having charge ‘q’ and mass ‘m’, when it is accelerated by a potential V. [CBSE (AI) 2013] Ans. de Broglie wavelength m= h = h p 2mqV p2 Hint: W = K = qV = 2m or p = 2mqV Q. 9. State de-Broglie hypothesis. [CBSE Delhi 2012] Ans. According to hypothesis of de Broglie “The atomic particles of matter moving with a given velocity, can display the wave like properties.” h i.e., m= mv (mathematically) Q. 10. The figure shows a plot of three curves a, b, c showing the variation of photocurrent vs. collector plate potential for three different intensities I1, I2, and I3 having frequencies n1, n2 and n3 respectively incident on a photosenitive surface. Point out the two curves for which the incident radiations have same frequency but different intensities. [CBSE Delhi 2009] Ans. Curves a and b have different intensities but same stopping potential, so curves ‘a’ and ‘b’ have same frequency but different intensities. Q . 11. The stopping potential in an experiment on photoelectric effect is 1.5 V. What is the maximum kinetic energy of the photoelectrons emitted? [CBSE (AI) 2009] Ans. Kmax = eVs = e (1.5V) = 1.5 eV = 1.5 × 1.6 × 10–19 J = 2.4 × 10–19 J Q . 12. The maximum kinetic energy of a photoelectron is 3 eV. What is its stopping potential? [CBSE (AI) 2009] Ans. (Ek)max = eVs Vs = (Ek) max = 3 eV =3 V Stopping potential, e e Q . 13. The stopping potential in an experiment on photoelectric effect is 2 V. What is the maximum kinetic energy of the photoelectrons emitted? [CBSE (AI) 2009] Ans. Maximum kinetic energy, (Ek)max =eVs       = e(2V)= 2 eV Q. 14. What is the stopping potential of a photocell, in which electrons with a maximum kinetic energy of 6 eV are emitted ? [CBSE (AI) 2008] Ans. Ek = eV0 ⇒ 6 eV = eV0 ⇒ V0 = 6 V The stopping potential V0= 6 volt (Negative). 462 Xam idea Physics–XII

Q. 15. The graph shows the variation of stopping Metal B Metal A potential with frequency of incident radiation 0' Frequency of incident for two photosensitive metals A and B. Which Stopping radiation ( ) potential one of the two has higher value of work-function? (V0) Justify your answer. [CBSE (AI) 2014] Ans. Metal A O 0 Since work function W = hν0 – We0 – We0 and ν'0 > ν0 so work function of metal A is more. Aliter: W l0 W0 e e On stopping potential axis – > – . Hence work function W′0 of metal A is more. Q . 16. An electron and a proton have the same kinetic energy. Which one of the two has the larger de Broglie wavelength and why? [CBSE (AI) 2012] Ans. An electron has the larger wavelength. h ? 1 for the same 2mEk m Reason: de-Broglie wavelength in terms of kinetic energy is m = kinetic energy. As an electron has a smaller mass than a proton, an electron has larger de Broglie wavelength than a proton for the same kinetic energy. 1 , where V is accelerating Q. 17. Plot a graph showing variation of de-Broglie wavelength λ versus V potential for two particles A and B carrying same charge but of masses m1, m2 (m1 > m2). Which one of the two represents a particle of smaller mass and why? [CBSE Delhi 2016] [HOTS] Ans. As, m = h or m=e h . 11 m2 2mqV 2q mo V m1 or m = h . 1 1/ V 1 2q m V As the charge on two particles is same, we get Slope ? 1 m Hence, particle with lower mass (m2) will have greater slope. Q . 18. Show the variation of photocurrent with collector plate potential for different frequencies but same intensity of incident radiation. [CBSE (F) 2011] [HOTS] Ans. Photo current ν1 ν1 >ν2 >ν3 ν 2ν3 Collector potential V1 V2 V3 Dual Nature of Matter and Radiation 463

Q. 19. (a) Draw a graph showing variation of photo-electric current (I) with anode potential (V) for different intensities of incident radiation. Name the characteristic of the incident radiation that is kept constant in this experiment. (b) If the potential difference used to accelerate electrons is doubled, by what factor does the de-Broglie wavelength associated with the electrons change? [CBSE (F) 2009] Ans. (a) The frequency of incident radiation was kept constant. (b) de-Broglie wavelength, m= h \\ 1 2mqV V If potential difference V is doubled, the de-Broglie wavelength is decreased to 1 times. v 2 Q . 20. (a) Define the term ‘intensity of radiation’ in photon picture. (b) Plot a graph showing the variation of photo current vs collector potential for three different intensities I1 > Iν2. > I3, two of which (I1 and I2) have the same frequency and the third has frequency ν1 > ν (c) Explain the nature of the curves on the basis of Einstein’s equation.  [CBSE South 2016] [HOTS] Ans. (a) The amount of light energy or photon energy incident per metre square per second is called intensity of radiation. (b)    ν2 = ν3 = ν Photo current I1 I2 I3 ν1 ν2 ν3 ν Collector potential (c) As per Einstein’s equation, (i) The stopping potential is same for I1 and I2 as they have the same frequency. (ii) The saturation currents are as shown in figure because I1 > I2 > I3. Q . 21. Show on a graph the variation of the de Broglie wavelength (λ) associated with an electron, with the square root of accelerating potential (V). [CBSE (F) 2012] [HOTS] Ans. We know m = 12.27 Ac V ∴ m V = constant V The nature of the graph between λ and V is hyperbola. Q. 22. Two metals A and B have work functions 4 eV and 10 eV respectively. Which metal has the higher threshold wavelength? Ans. Work function W = ho0 = hc m0 ⇒ m0 ? 1 W As WA < WB; (m0) A > (m0) B i.e., threshold wavelength of metal A is higher. 464 Xam idea Physics–XII

Q . 23. de Broglie wavelength associated with an electron accelerated through a potential difference V is l What will be the de Broglie wavelength when the accelerating potential is increased to 4V ? Ans. m 2 Reason: de Broglie wavelength associated with electron is m = h & m? 1 2meV V Obviously when accelerating potential becomes 4V, the de-Broglie wavelength reduces to half. Q. 24. (a) Draw a graph showing variation of photocurrent with anode potential for a particular intensity of incident radiation. Mark saturation current and stopping potential. (b) How much would stopping potential for a given photosensitive surface go up if the frequency of the incident radiations were to be increased from 4 × 1015 Hz to 8 × 1015 Hz? Ans. (a) Intercept of the graph with potential axis gives the stopping potential. (b) We have, hνin = eV ⇒ h (ν2 – ν1) TV = e = 6.62×10–34 × (8×1015 – 4×1015) 1.6×10–19 = 6.62×4×1015 ×10–34 V 1.6×10–19 = 16.55 V Q . 25. There are materials which absorb photons of shorter wavelength and emit photons of longer wavelength. Can there be stable substances which absorb photons of larger wavelength and emit light of shorter wavelength? [NCERT Exemplar] Ans. In the first case, energy given out is less than the energy supplied. In the second case, the material has to supply the energy as the emitted photon has more energy. This cannot happen for stable substances. Q. 26. Do all the electrons that absorb a photon come out as photoelectrons? [NCERT Exemplar] [HOTS] Ans. No, most electrons get scattered into the metal. Only a few come out of the surface of the metal. Q . 27. Electrons are emitted from a photosensitive surface when it is illuminated by green light but electron emission does not take place by yellow light. Will the electrons be emitted when the surface is illuminated by (i) red light, and (ii) blue light? [HOTS] Ans. (i) No (ii) Yes. Reason: According to colour sequence VIBGYOR, the frequency of red light photons is less than threshold frequency of photometal but frequency of blue light photons is more than threshold frequency of photometal; so (i) electrons will not be emitted by red light and (ii) electrons will be emitted by blue light. Dual Nature of Matter and Radiation 465

Q. 28. In a photoelectric effect, the yellow light is just able to emit electrons, will green light emit photoelectrons? What about red light? [HOTS] Ans. Energy of photon E= hc ? 1 m m As λgreen <λyellow so green light photon has more energy than yellow light photon, so green light will eject electron. As λred > λyellow so red light photon has lesser energy than yellow light photon, so red light will not be able to eject electrons. Q. 29. Work function of aluminium is 4.2 eV. If two photons, each of energy 2.5 eV, are incident on its surface, will the emission of electrons take place? Justify your answer. [HOTS] Ans. In photoelectric effect, a single photon interacts with a single electron. As individual photon has energy (2.5 eV) which is less than work function, hence emission of electron will not take place. Short Answer Questions–I [2 marks] Q. 1. Write Einstein’s photoelectric equation and point out any two characteristic properties of photons on which this equation is based. [CBSE (AI) 2013] Ans. If radiation of frequency (ν) greater than threshold frequency (ν0) irradiate the metal surface, electrons are emitted out from the metal. So Einstein’s photoelectric equation can be given as Kmax = 1 mvm2 ax = hν – hν0 2 Characteristic properties of photons: (i) Energy of photon is directly proportional to the frequency (or inversely proportional to the wavelength). (ii) In photon-electron collision, total energy and momentum of the system of two constituents remains constant. (iii) In the interaction of photons with the free electrons, the entire energy of photon is absorbed. Q. 2. Write three characteristic features in photoelectric effect which cannot be explained on the basis of wave theory of light, but can be explained only using Einstein’s equation. [CBSE Delhi 2016] Ans. The three characteristic features which cannot be explained by wave theory are: (i) Kinetic energy of emitted electrons is found to be independent of the intensity of incident light. (ii) There is no emission of electrons if frequency of incident light is below a certain frequency (threshold frequency). (iii) Photoelectric effect is an instantaneous process. Q. 3. A proton and an electron have same velocity. Which one has greater de Broglie wavelength and why? [CBSE (AI) 2012] Ans. de Broglie wavelength (λ) is given as m= h mv Given vp = ve where vp= velocity of proton and ve = velocity of electron Since mp > me From the given relation m ? 1 , hence λp < λe m Thus, electron has greater de Broglie wavelength, if accelerated with same speed. 466 Xam idea Physics–XII

Q. 4. What is meant by work function of a metal? How does the value of work function influence the kinetic energy of electrons liberated during photoelectron emission? [CBSE Delhi 2013; (AI) 2013] Ans. Work Function: The minimum energy required to free an electron from metallic surface is called the work function. Smaller the work function, larger the kinetic energy of emitted electron. Q. 5. Monochromatic light of frequency 6 × 1014 Hz is produced by a laser. The power emitted is 2.0 × 10–3 W. How many photons per second on an average are emitted by the source? [CBSE Guwahati 2015] nhν Ans. Power of radiation, P= t = Nhν, where N is number of photons per sec. or N= P hν 2.0 # 10–3 = 6.63 # 10–34 # 6 # 1014 = 5 × 1015 photons per second Q. 6. Plot a graph showing the variation of photoelectric current with intensity of light. The work function for the following metals is given: Na: 2.75 eV and Mo: 4.175 eV. Which of these will not give photoelectron emission from a radiation of wavelength 3300 Å from a laser beam? What happens if the source of laser beam is brought closer? [CBSE (F) 2016] Ans. Energy of photon E= hc Joule m Photoelectric current = hc eV em 6.63 # 10–34 # 3 # 108 = 1.6 # 10–19 # 3.3 # 10–7 eV = 3.76 eV Since W0 of Mo is greater than E, ∴ Mo will not give photoemission. There will be no effect of bringing source closer in the case of Mo. In case of Na, photocurrent will increase. Q. 7. The given graph shows the variation of photo-electric current (I) Intensity of light with the applied voltage (V) for two different materials and for two different intensities of the incident radiations. Identify and explain using Einstein’s photo electric equation for the pair of curves that correspond to (i) different materials but same intensity of incident radiation, (ii) different intensities but same materials. [CBSE East 2016] I 1 2 3 4 V Ans. (a) 1 and 2 correspond to same intensity but different material. (b) 3 and 4 correspond to same intensity but different material. This is because the saturation currents are same and stopping potentials are different. Dual Nature of Matter and Radiation 467

(a) 1 and 3 correspond to different intensity but same material. (b) 2 and 4 correspond to different intensity but same material. This is because the stopping potentials are same but saturation currents are different. Q. 8. Plot a graph showing the variation of stopping potential with the 21 frequency of incident radiation for two different photosensitive Vs materials having work functions W1 and W2 (W1>W2). On what θθ factors does the (i) slope and (ii) intercept of the lines depend? [CBSE Delhi 2010] Ans. The graph of stopping potential Vs and frequency (ν) for two 2 (W2/e ) photosensitive materials 1 and 2 is shown in fig. (i) Slope of graph tan i= h = universal constant. 1 (W1/e ) e (ii) Intercept of lines depend on the work function. Q. 9. An electron is accelerated through a potential difference of 100 V. What is the de Broglie wavelength associated with it? To which part of the electromagnetic spectrum does this value of wavelength correspond? [CBSE Delhi 2010] Ans. de Broglie wavelength, md= h n = h p 2meV = 6.63 # 10–34 2 # 9.1 # 10–31 # 1.6 # 10–19 # 100 = 1.227×10–10 m = 1.227 Å This wavelength belongs to X-ray spectrum. Q . 10. An electromagnetic wave of wavelength m1 is incident on a photosensitive surface of negligible work function. If the photo-electrons emitted from this surface have the de-Broglie wavelength prove that m = d 2mc nm12 [CBSE Delhi 2008] h Ans. Kinetic energy of electrons, Ek = energy of photon of EM wave = hc …(i) m de Broglie wavelength, m1 = h or m12 = h2 Using (i), we get 2mEk 2mEk m12 = h2 & m = d 2mc nm12 h 2md hc n m Q. 11. An α-particle and a proton of the same kinetic energy are in turn allowed to pass through a magnetic field B , acting normal to the direction of motion of the particles. Calculate the ratio of radii of the circular paths described by them. [CBSE 2019 (55/1/1)] Ans. Radius r = mv = 2mK qB qB Ka = Kproton Ma = 4 MP qa = 2qP 468 Xam idea Physics–XII

ra = 2ma K rp qa B 2mp K qpB = ma × qp mp qa = 4 × 1 = 1 2 Q . 12. There are two sources of light, each emitting with a power 100W. One emits X-rays of wavelength 1 nm and the other visible light at 500 nm. Find the ratio of number of photons of X-rays the photons of visible light of the given wavelength. [NCERT Exemplar] Ans. Total E is constant. Let n1 and n2 be the number of photons of X-rays and visible region. n1 E1 = n2 E2 & n1 hc = n2 hc m1 m2 nn12 = m1 & n1 = 1 Electron m2 n2 500 Q. 13. Consider Fig. for photoemission. How would you reconcile with momentum-conservation? No light (Photons) have momentum in a different direction than the emitted electrons. [NCERT Exemplar] Ans. The momentum is transferred to the metal. At the microscopic level, atoms Light absorb the photon and its momentum is transferred mainly to the nucleus and Metal electrons. The excited electron is emitted. Conservation of momentum needs to be accounted for the momentum transferred to the nucleus and electrons. Q . 14. A photon and a proton have the same de-Broglie wavelength λ. Prove that the energy of the photon is (2mλc/h) times the kinetic energy of the proton. [CBSE 2019 (55/2/1)] Ans. Energy of photon EP = hc m For proton m = h mv h mv = m Kinetic energy of proton Ek = 1 mv2 2 h2 Ek = 1 mm2 2 EP = d 2mmc nEk h Q. 15. If light of wavelength 412·5 nm is incident on each of the metals given below, which ones will show photoelectric emission and why? [CBSE 2018] Metal Work Function (eV) Na 1.92 K 2.15 Ca 3.20 Mo 4.17 Dual Nature of Matter and Radiation 469

Ans. The energy of the incident photon, E = ho = hc m = 6.63×10–34 ×3×108 J 412.5 ×10 –9 = 0.048 ×10 –17 eV =3 eV 1.6 ×10–19 Metals having work function less than energy of the incident photon will show photoelectric effect. Hence, only Na and K will show photoelectric emission. Short Answer Questions–II [3 marks] Q. 1. Explain briefly the reasons why wave theory of light is not able to explain the observed features of photo-electric effect. [CBSE Delhi 2013; (AI) 2013; (F) 2010; 2019 (55/2/1)] Ans. The observed characteristics of photoelectric effect could not be explained on the basis of wave theory of light due to the following reasons. (i) According to wave theory, the light propagates in the form of wavefronts and the energy is distributed uniformly over the wavefronts. With increase of intensity of light, the amplitude of waves and the energy stored by waves will increase. These waves will then, provide more energy to electrons of metal; consequently, the energy of electrons will increase. Thus, according to wave theory, the kinetic energy of photoelectrons must depend on the intensity of incident light; but according to experimental observations, the kinetic energy of photoelectrons does not depend on the intensity of incident light. (ii) According to wave theory, the light of any frequency can emit electrons from metallic surface provided the intensity of light be sufficient to provide necessary energy for emission of electrons, but according to experimental observations, the light of frequency less than threshold frequency cannot emit electrons; whatever the intensity of incident light may be. (iii) According to wave theory, the energy transferred by light waves will not go to a particular electron, but it will be distributed uniformly to all electrons present in the illuminated surface. Therefore, electrons will take some time to collect the necessary energy for their emission. The time for emission will be more for light of less intensity and vice versa. But experimental observations show that the emission of electrons take place instantaneously after the light is incident on the metal; whatever the intensity of light may be. Q. 2. Write Einstein’s photoelectric equation. State clearly the three salient features observed in photoelectric effect which can explain on the basis of this equation. The maximum kinetic energy of the photoelectrons gets doubled when the wavelength of light incident on the surface changes ftrhoemmemt1atlosmu2rf.aDcee.r ive the expressions for the threshold wavelength m0 and work function for [CBSE Delhi 2015; (AI) 2010] Ans. Einstein’s photoelectric equation: hν = hν0 + eV0 where ν = incident frequency, ν0 = threshold frequency, V0 = stopping potential (i) Incident energy of photon is used in two ways (a) to liberate electron from the metal surface (b) rest of the energy appears as maximum energy of electron. (ii) Only one electron can absorb energy of one photon. Hence increasing intensity increases the number of electrons hence current. (iii) If incident energy is less than work function, no emission of electron will take place. (iv) Increasing ν (incident frequency) will increase maximum kinetic energy of electrons but number of electrons emitted will remain same. 470 Xam idea Physics–XII

For wavelength λ1 hc m1 = z0 + K = z0 + eV0 …(i) where K= eV0 For wavelength λ2 hc m2 = z0 + 2eV0 …(ii) (because KE is doubled) From equation (i) and (ii), we get hc = z0 + 2 e hc – z0o = z0 + 2hc – 2z0 m2 m1 m1 ⇒ z0 = 2hc – hc m1 m2 For threshold wavelength λ0 kinetic energy, K = 0, and work function z0 = hc m0 hc 2hc hc ∴ m0 = m1 – m2 ⇒ 1 = 2 – 1 & m0 = m1 m2 m0 m1 m2 2m2 – m1 Work function, z0 = hc (2m2 – m1) m1 m2 Q. 3. Using photon picture of light, show how Einstein’s photoelectric equation can be established. Write two features of photoelectric effect which cannot be explained by wave theory. [CBSE (AI) 2017] Ans. In the photon picture, energy of the light is assumed to be in the form of photons each carrying energy. When a photon of energy ‘hν’ falls on a metal surface, the energy of the photon is absorbed by the electrons and is used in the following two ways: (i) A part of energy is used to overcome the surface barrier and come out of the metal surface. This part of energy is known as a work function and is expressed as φ0 = hν0. (ii) The remaining part of energy is used in giving a velocity ‘v’ to the emitted photoelectron 1 which is equal to the maximum kinetic energy of photo electrons c 2 mvm2 ax m . (iii) According to the law of conservation of energy, ho = z0 + 1 mvm2 ax 2 ⇒ ho = ho0 + 1 mvm2 ax & ho = ho0 + KEmax 2 ⇒ KEmax = hν – hν0 or KEmax = hν – φ0 This equation is called Einstein photoelectric equation. Features which cannot be explained by wave theory: (i) The process of photoelectric emission is instantaneous in nature. (ii) There exists a ‘threshold frequency’ for each photosensitive material. (iii) Maximum kinetic energy of emitted electrons is independent of the intensity of incident light. Q. 4. A proton and an alpha particle are accelerated through the same potential. Which one of the two has (i) greater value of de Broglie wavelength associated with it and (ii) less kinetic energy? Give reasons to justify your answer. [CBSE North 2016, Delhi 2014] Dual Nature of Matter and Radiation 471

Ans. (i) de Broglie wavelength m= h = h p 2mqV For same V, m a 1 mq mp = ma qa = 4m p . 2e ma mpqp mp e = 8=2 2 Clearly, λp >λα. Hence, proton has a greater de-Broglie wavelength. (ii) Kinetic energy, K =qV For same V, K α q qp Kp qa Ka = = e = 1 2e 2 Clearly, Kp < Kα. Hence, proton has less kinetic energy. Q. 5. Define the terms (i) ‘cut-off voltage’ and (ii) ‘threshold frequency’ in relation to the phenomenon of photoelectric effect. Using Einstein’s photoelectric equation show how the cut-off voltage and threshold frequency for a given photosensitive material can be determined with the help of a suitable plot/graph. [CBSE (AI) 2012] Ans. (i) Cut off or stopping potential is that minimum value of negative potential at anode which just stops the photo electric current. Y (ii) For a given material, there is a minimum frequency of light below which no photo electric emission will take place, this frequency is called as threshold frequency. V0 h e By Einstein’s photo electric equation Slope = KEmax = hc – z = hν – hν0 X m eV0 = hν – hν0 Intercept = – /e 0 V0 = h ν – h e e ν0 Clearly, V0 – ν graph is a straight line. Q. 6. Write two characteristic features observed in photoelectric effect which support the photon picture of electromagnetic radiation. Draw a graph between the frequency of incident radiation K.E. of electron (eV) (n) and the maximum kinetic energy of the electrons emitted from the surface of a photosensitive material. State clearly how this graph can be used to determine (i) Planck’s constant and (ii) work function of the material. [CBSE Delhi 2017, (F) 2012] Ans. (a) All photons of light of a particular frequency ‘ν’ have X same energy and momentum whatever the intensity of 0 (Hz) radiation may be. (b) Photons are electrically neutral and are not affected by presence of electric and magnetic fields, 472 Xam idea Physics–XII

(i) From this graph, the Planck constant can be calculated by the slope of the current h = T (KE) Tν (ii) W ork function is the minimum energy required to eject the photo-electron from the metal surface. φ = hν0, where ν0 = Threshold frequency Q. 7. Sketch the graphs showing variation of stopping potential with frequency of incident radiations for two photosensitive materials A and B having threshold frequencies νA > νB. (i) In which case is the stopping potential more and why? (ii) Does the slope of the graph depend on the nature of the material used? Explain. [CBSE Central 2016] Ans. (i) From the graph for the same value of ‘ν’, stopping potential is more for material ‘B’. From Einstein’s photoelectric equation V BA eV0 = hn – hn0 h Stopping h h e Potential V0 = e o – e o0 = (o – o0) Q. 8 . A(iip) rN∴otoo,naVsa0nslidosphaeidgihseuegrtievfreoonrnbloaywreeheracvwcaheluilceehroaitfseνad0uthnrivoeurgsahl constant. accelerating vB vA v one of the two has the same potential. Which (i) greater value of de-Broglie wavelength associated with it, and (ii) less momentum? Give reasons to justify your answer. [CBSE Delhi 2014] Ans. (i) de Broglie wavelength, m = h 2mqV Here V is same for proton and deutron. As mass of proton < mass of deutron and qp = qd Therefore, λp > λd for same accelerating potential. (ii) We know that momentum = h m Therefore, λp > λd So, momentum of proton will be less than that of deutron. Q. 9. A beam of monochromatic radiation is incident on a photosensitive surface. Answer the following questions giving reasons: (i) Do the emitted photoelectrons have the same kinetic energy? (ii) Does the kinetic energy of the emitted electrons depend on the intensity of incident radiation? (iii) On what factors does the number of emitted photoelectrons depend? [CBSE (F) 2015] Ans. In photoelectric effect, an electron absorbs a quantum of energy hν of radiation, which exceeds the work function, an electron is emitted with maximum kinetic energy, Kmax = hν – W (i) No, all electrons are bound with different forces in different layers of the metal. So, more tightly bound electron will emerge with less kinetic energy. Hence, all electrons do not have same kinetic energy. (ii) No, because an electron cannot emit out if quantum energy hν is less than the work function of the metal. The K.E. depends on energy of each photon. (iii) Number of emitted photoelectrons depends on the intensity of the radiations provided the quantum energy hν is greater than the work function of the metal. Dual Nature of Matter and Radiation 473

Q. 10. Why are de Broglie waves associated with a moving football not visible? The wavelength ‘ m ’ of a photon and the de Broglie wavelength of an electron have the same value. Show that the energy of photon is 2mmc times the kinetic energy of electron, where m, h c, h have their usual meanings. [CBSE (F) 2016] Ans. Due to large mass of a football the de Broglie wavelength associated with a moving football is much smaller than its dimensions, so its wave nature is not visible. h h de Broglie wavelength of electron m = mv &v= mm ....(i) energy of photon E = hc (because λ is same) ...(ii) m Ratio of energy of photon and kinetic energy of electrons E = hc/m = 2hc Ek mmv2 1 mv2 2 Substituting value of v from (i), we get E = 2hc = 2mmc Ek mm (h/mm)2 h ∴ Energy of photon = 2mmc × kinetic energy of electron h Q. 11. An α-particle and a proton are accelerated from rest by the same potential. Find the ratio of their de- Broglie wavelengths. [CBSE Delhi 2017, (AI) 2010] Ans. de-Broglie wavelength m = h= h 2mE 2mqV h For -particle, ma = For proton, mp = 2ma qa V h 2mp qp V ∴ ma = mpqp But mp ma qa ma = 4, qa = 2 mp qp ∴ ma = 1 .1 = 1 = 1 . mp 4 2 8 22 Q . 12. A proton and an α-particle have the same de-Broglie wavelength. Determine the ratio of (i) their accelerating potentials (ii) their speeds. [CBSE Delhi 2015; 2019 (55/4/1)] Ans. de Broglie wavelength m= h = h p 2mqV where, m = mass of charge particle, q = charge of particle, V = potential difference (i) m2 = h2 & V = h2 ∴ 2mqV 2mqm2 Vp = 2ma qa = 2 # 4m2q = 8 Va 2mp qp 2mq 1 ∴ Vp : Vα = 8 : 1 474 Xam idea Physics–XII

(ii) m = h , m p = m h , ma = h mv pv ma va p m p = ma & h = h mpvp ma va vp = ma = 4 = 4 :1 va mp 1 Q. 13. An electron and a proton, each have de Broglie wavelength of 1.00 nm. (a) Find the ratio of their momenta. (b) Compare the kinetic energy of the proton with that of the electron. [CBSE (F) 2013] h h Ans. (a) me = pe and m p = pp ,me = m p = 1.00 nm So, me = pp = 1 & pp = 1 = 1:1 mp pe 1 pe 1 (b) From relation K= 1 mv2 = p2 2 2m 2 Ke = pe2 and Kp = p p 2me 2m p Kp = p 2 2me = me Ke p pe2 mp # 2m p Since me <<< mp. So Kp <<< Ke. Kp = 9.1 # 10–31 = 5.4 # 10–4 Ke 1.67 # 10–27 Q . 14. Write briefly the underlying principle used in Davison-Germer experiment to verify wave nature of electrons experimentally. What is the de-Broglie wavelength of an electron with kinetic energy (KE) 120 eV? [CBSE South 2016] Ans. Principle: Diffraction effects are observed for beams of electrons scattered by the crystals. m= h = h= h p 2mEk 2meV = 6.63 # 10–34 2 # 9.1 # 10–31 # 1.6 # 10–19 # 120 λ = 0.112 nm Q. 15. (a) Define the term ‘intensity of radiation’ in terms of photon picture of light. (b) Two monochromatic beams, one red and the other blue, have the same intensity. In which case (i) the number of photons per unit area per second is larger, (ii) the maximum kinetic energy of the photoelectrons is more? Justify your answer. [CBSE Patna 2015] Ans. (a) The number of photons incident normally per unit area per unit time is determined the intensity of radiations. (b) (i) Red light, because the energy of red light is less than that of blue light (hν)R < (hν)B (ii) Blue light, because the energy of blue light is greater than that of red light (hν)B > (hν) R Q 16. Determine the value of the de Broglie wavelength associated with the electron orbiting in the ground state of hydrogen atom (Given En = – (13.6/n2) eV and Bohr radius r0 = 0.53 Å). How will the de Broglie wavelength change when it is in the first excited state? [CBSE Bhubaneshwar 2015] Dual Nature of Matter and Radiation 475

Ans. In ground state, the kinetic energy of the electron is K = –E = + 13.6 eV = 13.6 # 1.6 # 10–19 J = 2.18×10–18 J 12 de Broglie wavelength, m= h = h p 2mK m1 = 6.63 # 10–34 2 # 9.1 # 10–31 # 2.18 # 10–18 = 0.33×10–9 m = 0.33 nm Kinetic energy in the first excited state (n=2) K = – E = + 13.6 eV =+ 3.4 eV = 3.4×1.6×10–19 J = 0.54×10–18 J 22 de Broglie wavelength, m2 = h 2mK = 6.63 # 10–34 2 # 9.1 # 10–31 # 0.544 # 10–18 = 2 × 0.33 nm = 0.66 nm i.e., de Broglie wavelength will increase (or double). Q 17. Define the term ‘intensity of radiation’ in photon picture of light. Ultraviolet light of wavelength 2270 Å from 100 W mercury source irradiates a photo cell made of a given metal. If the stopping potential is –1.3 V, estimate the work function of the metal. How would the photocell respond to a high intensity (~ 105 Wm–2) red light of wavelength 6300 Å produced by a laser? [CBSE Bhubaneshwar 2015] Ans. The intensity of light of certain frequency (or wavelength) is defined as the number if photons passing through unit area in unit time. For a given wavelength, (λ) of light hc = W + K m = W + eVs (where Vs is stopping potential) 6.63×10–34 ×3×108 = W + 1.6×10–19 × (–1.3 eV) 2270×10–10 ∴ W = f 6.63×10–34 ×3×108 –1.3 p eV 2270×10–10 ×1.6×10–19 W = 4.2 eV The wavelength of red light 6300 Å >> 2270 Å. So, the energy of red light must be E = ho = hc in eV em = 6.63×10–34 ×3×108 1.6×10–19 ×6300×10–10 = 6.63×3 ×10 = 198.9 eV = 1.973 eV 1.6×63 1.6×63 The energy of red light is very less than its work function, even intensity is very high. Hence no emission of electron is possible. 476 Xam idea Physics–XII

Q . 18. In the study of a photoelectric effect the graph between the stopping potential V and frequency n of the incident radiation on two different metals P and Q is shown below: (i) Which one of the two metals has higher threshold frequency? (ii) Determine the work function of the metal which has greater value. (iii) Find the maximum kinetic energy of electron emitted by light of frequency 8 × 1014 Hz for this metal. [CBSE Delhi 2017] Ans. (i) Threshold frequency of P is 3 × 1014 Hz. Threshold frequency of Q is 6 × 1014 Hz. Clearly Q has higher threshold frequency. (ii) Work function of metal Q, f0 = hn0 = (6.6 × 10–34) × 6 × 1014 J = 39.6 ×10–20 eV = 2.5 eV 1.6 ×10–19 (iii) Maximum kinetic energy, Kmax = hn – hn0 = h(n – n0) = 6.6 × 10–34 (8 × 1014 – 6 × 1014) = 6.6 × 10–34 × 2 × 1014 J = 6.6 ×10–34 × 2×1014 eV 1.6 ×10–19 ∴ Kmax = 0.83 eV Q. 19. Two monochromatic beams A and B of equal intensity I, hit a screen. The number of photons hitting the screen by beam A is twice that by beam B. Then what inference can you make about their frequencies? [NCERT Exemplar] Ans. Let no. of photons falling per second of beam A = nA No. of photons falling per second of beam B = nB Energy of beam A = hνA Energy of beam B =hνB According to question, I = nA hνA = nB hνB nA 2nB nB = νB or, nB = νB & νB = 2ν A νA νA The frequency of beam B is twice that of A. Q . 20. A monochromatic light source of power 5 mW emits 8 × 1015 photons per second. This light ejects photo electrons from a metal surface. The stopping potential for this set up is 2 eV. Calculate the work function of the metal. [CBSE Sample Paper 2016] Ans. P = 5×10–3 W, N = 8 × 1015 photons per second Energy of each photon, 6.25 # 10–19 5 # 10–3 1.6 # 10–19 E = P = 8 # 1015 = 6.25 # 10–19 J = eV N Dual Nature of Matter and Radiation 477

E = 3.9 eV Work function, W0 = E – V0 = (3.9 – 2) eV= 1.9 eV Long Answer Questions [5 marks] Q. 1. Describe an experimental arrangement to study photoelectric effect. Explain the effect of (i) intensity of light on photoelectric current, (ii) potential on photoelectric current and (iii) frequency of incident radiation on stopping potential. Ans. Experimental study of Photoelectric Effect: The apparatus consist of an evacuated glass or quartz tube which encloses a photosensitive plate C (called emitter) and a metal plate A (called collector). A transparent window W is sealed on the glass tube which can be covered with a filter for a light of particular radiation. This will allow the light of particular wavelength to pass through it. The plate A can be given a desired positive or negative potential with respect to plate C, using the arrangement as shown in figure. S Source Quartz window Evacuated glass tube Photosensitive W plate Collector e e e A C e e µA e e Reversing key _+ K Battery Working: When a monochromatic radiations of suitable frequency obtained from source S fall on the photosensitive plate C, the photoelectrons are emitted from C, which gets accelerated towards the plate A (collector) if it is kept at positive potential. These electrons flow in the outer circuit resulting in the photoelectric current. Due to it, the microammeter shows a deflection. The reading of microammeter measures the photoelectric current. This experimental arrangement can be used to study the variation of photoelectric current with the following quantities. (i) Effect of intensity of the incident radiation: By varying P the intensity of the incident radiations, keeping the Photoelectric frequency constant, it is found that the photoelectric current current varies linearly with the intensity of the incident radiation. Also, the number of photoelectrons emitted per second O intensity is directly proportional to the intensity of the incident radiations. 478 Xam idea Physics–XII

(ii) Effect of potential of plate A w.r.t place C: It is found that the photoelectric current increases gradually with the increase in positive potential of plate A. At one stage for a certain positive potential of plate A, the photoelectric current becomes maximum or saturates. After this if we increase the positive potential of plate A, there will be Photoelectric I3 no increase in the photoelectric current. Current This maximum value of current is called I2 I1 saturation current: The saturation current I3> I2>I1 corresponds to the state when all the photoelectrons emitted from C reach the plate A. –V° Potential Now apply a negative potential on plate A w.r.t. plate C. We will note that the photoelectric current decreases, because the photoelectrons emitted from C are repelled and only energetic photoelectrons are reaching the plate A. By increasing the negative potential of plate A, the photoelectric current decreases rapidly and becomes zero at a certain value of negative potential V0 on plate A. This maximum negative potential V0, given to the plate A w.r.t. plate C at which the photoelectric current becomes zero is called stopping potential or cut off potential. Kmax = eV0 = 1 m Vm2 ax 2 where e = charge on electron, m = mass of electrons Vmax = maximum velocity of emitted photoelectrons. The value of stopping potential is independent of the intensity of the incident radiation. It means, the maximum kinetic energy of emitted photoelectrons depends on the radiation source and nature of material of Photoelectric plate C but is independent of the intensity of Current ν3 Saturation ν2 Current incident radiation. ν1 Potential (iii) Effect of frequency of the incident radiation: When we take the radiations of different frequencies but of same intensity, then the value of stopping potential is different for ν°3 ν°2 ν°1 radiation of different frequency. The value of stopping potential is more negative for radiation of higher incident frequency. The value of saturation current depends on the intensity of incident radiation but is independent of the frequency of the incident radiation. (iv) Effect of frequency on stopping potential: For a given Stopping Metal A photosensitive material, the stopping potential varies Potential Metal B linearly with the frequency of the incident radiation. For every photosensitive material, there is a certain minimum cut off frequency ν0 (threshold frequency) for which the stopping potential is zero. The intercept on the potential axis = – z0 = – hν0 . O ν°1 ν°2 Frequency of e e -φ°1 Radiation e Hence, work function φ0 = e × magnitude of intercept -φ°2 on the potential axis e Dual Nature of Matter and Radiation 479

Q. 2. Derive Einstein’s photoelectric equation 1 mv2 = hν – hν0 . 2 Ans. Einstein’s Explanation of Photoelectric Effect: Einstein’s Photoelectric Equation Einstein explained photoelectric effect on the basis of quantum theory. The main points are Photoelectron 1. Light is propagated in the form of bundles of energy. Each bundle of IncPidheontot n hν EK energy is called a quantum or photon and has energy hν where h = Planck’s constant and ν = frequency of light. 2. The photoelectric effect is due to collision of a photon of incident light w and a bound electron of the metallic cathode. Metal 3. When a photon of incident light falls on the metallic surface, it is completely absorbed. Before being absorbed it penetrates through a distance of nearly 10–8 m (or 100 Å). The absorbed photon transfers its whole energy to a single electron. The energy of photon goes in two parts: a part of energy is used in releasing the electron from the metal surface (i.e., in overcoming work function) and the remaining part appears in the form of kinetic energy of the same electron. If ν be the frequency of incident light, the energy of photon = hν. If W be the work function of metal and EK the maximum kinetic energy of photoelectron, then according to Einstein’s explanation. hν = W + EK or EK = hν – W ...(i) This is called Einstein’s photoelectric equation. If ν0 be the threshold frequency, then if frequency of incident light is less then ν0 no electron will be emitted and if the frequency of incident light be ν0 then EK = 0; so from equation (i) 0 = hν0 – W or W = hν0 c m0 If λ0 be the threshold wavelength, then ν0 = , where c is the speed of light in vacuum ∴ Work function inWeq=uahtiνo0n=(i)mh,c0w e get ...(ii) Substituting this value EK = hν – hν0 & 1 mv2 = hν – hν0 ...(iii) 2 This is another form of Einstein’s photoelectric equation. Q. 3. (a) Give a brief description of the basic elementary process involved in the photoelectric emission in Einstein’s picture. (b) When a photosensitive material is irradiated with the light of frequency n, the maximum speed of electrons is given by vmax. A plot of v2max is found to vary with frequency ν as shown in the figure. Use Einstein’s photoelectric equation to find the expressions for (i) Planck’s constant and v2max n T (ii) work function of the given 0 v photosensitive material, in terms l of the parameters l, n and mass m of the electron. Ans. (a) Refer to Q. 2 above. (b) (i) v12 and v22 are the velocities of the emitted electrons for radiations of frequencies n1 > n and n2 > n respectively. So, 480 Xam idea Physics–XII

hν1 = hν + 1 mv12 …(i) and 2 hν2 = hν + 1 mv22 …(ii) 2 From equation (i) and (ii), we get h (ν2 – ν1) = 1 m (v22 – v12) 2 h= 1 m (v22 – v12) 2 ∴ (ν2 – ν1) Slope of vm2 ax vs frequency graph is v22 v12 tan i = (ν2 – ν1) – ∴ h = 1 m . tan i 2 From graph tan i = l n 1 l So, h = 2 mc n m …(iii) (ii) From graph, the work function of the material is     W = hn …(iv) From equations (iii) and (iv), we get W = 1 mc l m× n = 1 ml 2 n 2 Q. 4. Describe Davisson and Germer’s experiment to demonstrate the wave nature of electrons. Draw a labelled diagram of apparatus used. [CBSE (F) 2014] Ans. Davisson and Germer Experiment: In 1927 Electron Davisson and Germer performed a diffraction gun experiment with electron beam in analogy with X -ray diffraction to observe the wave nature of matter. Apparatus: It consists of three parts: (i) Electron Gun: It gives a fine beam of electrons. Incident Scattered beam Electron de Broglie used electron beam of energy 54 eV. beam detector de Broglie wavelength associated with this beam m= h Nickel 2mEK Crystal Here m = mass of electron = 9.1×10–31 kg EK = Kinetic energy of electron = 54 eV = 54×1.6×10–19 joule = 86.4×10–19 joule ∴ m= 6.6 # 10–34 2 # 9.1 # 10–31 # 86.4 # 10–19 = 1.66×10–10 m = 1.66 Å (ii) Nickel Crystal: The electron beam was directed on nickel crystal against its (iii) face. The smallest separation between nickel atoms is 0.914 Å. Nickel crystal behaves as diffraction grating. Dual Nature of Matter and Radiation 481

(iii) Electron Detector: It measures the intensity of electron beam diffracted from nickel crystal. It may be an ionisation chamber fitted with a sensitive galvanometer. The energy of electron beam, the angle of incidence of beam on nickel crystal and the position of detector can all be varied. Method: The crystal is rotated in small steps to change the angle (α say) between incidence and scattered directions and the corresponding intensity (I) of scattered beam is measured. The variation of the intensity (I) of the scattered electrons with the angle of scattering α is obtained for different accelerating voltages. The experiment was performed by varying the accelerating 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 accelerating voltage of 54 V at a scattering angle α = 50° ∴ From Bragg’s law 2d sin θ = nλ 25o 50o Here n = 1, d = 0.914 Å, θ = 65° θ = 65o ∴ m = 2d sin i Nickel Crystal n = 2 # (0.914 Ac ) sin 65° 1 = 2 # 0.914 # 0.9063 Ac = 1.65 Ac The measured wavelength is in close agreement with the estimated de Broglie wavelength. Thus the wave nature of electron is verified. Later on G.P. Thomson demonstrated the wave nature of fast electrons. Due to their work Davission and G.P. Thomson were awarded Nobel prize in 1937. Later on experiments showed that not only electrons but all material particles in motion (e.g., neutrons, α-particles, protons etc.) show wave nature. Self-Assessment Test Time allowed: 1 hour Max. marks: 30 1. Choose and write the correct option in the following questions. (3 × 1 = 3) (i) Photoelectric emission occurs only when the incident light has more than a certain minimum (a) power (b) wavelength (c) intensity (d) frequency (ii) The threshold frequency for a photosensitive metal is 3.3 × 1014 Hz. If light of frequency 8.2 × 1014 Hz is incident on this metal, the cut-off voltage for the photoelectron emission is nearly (a) 1 V (b) 2 V (c) 3 V (d) 5 V (iii) When the light of frequency 2n0 (where n0 is threshold frequency), is incident on a metal plate, the maximum velocity of electrons emitted is v1. When the frequency of the incident radiation is increased to 5n0, the maximum velocity of electrons emitted from the same plate is v2. The ratio of v1 to v2 is (a) 1 : 2 (b) 1 : 4 (c) 4 : 1 (d) 2 : 1 482 Xam idea Physics–XII

2. Fill in the blanks. (2 × 1 = 2) (i) The maximum kinetic energy of emitted photoelectrons is independent of ______________ of incident radiation. (ii) The expression for de Broglie wavelength of an electron moving under a potential difference of V volts is ______________. 3. Plot a graph of the de-Broglie wavelength associated with a proton versus its momentum. 1 4. Draw graphs showing variation of photoelectric current with applied voltage for two incident radiations of equal frequency and different intensities. Mark the graph for the radiation of higher intensity. 1 5. Show on a graph variation of the de Broglie wavelength (λ) associated with the electron versus 1/ V , where V is the accelerating potential for the electron. 1 6. A deuteron and an alpha particle are accelerated with the same accelerating potential. Which one of the two has (a) greater value of de Broglie wavelength, associated with it? and (b) less kinetic energy? Explain. 2 7. The work function of caesium is 2.14 eV. Find (i) the threshold frequency for caesium and (ii) wavelength of incident light if the photocurrent is brought to zero by a stopping potential of 0.60 V. 2 8. Light of wavelength 2500 Å falls on a metal surface of work function 3.5 V. What is the kinetic energy (in eV) of (i) the fastest and (ii) the slowest electronic emitted from the surface? If the same light falls on another surface of work function 5.5 eV, what will be the energy of emitted electrons? 2 9. Plot a graph showing variation of de Broglie wavelength (λ) associated with a charged particle of mass m, versus 1/ V , where V is the potential difference through which the particle is accelerated. How does this graph give us the information regarding the magnitude of the charge of the particle?  2 10. Deduce de Broglie wavelength of electrons accelerated by a potential of V volt. Draw a schematic diagram of a localized wave describing the wave nature of moving electron. 3 11. When a given photosensitive material is irradiated with light of frequency n, the maximum speed of the emitted photoelectrons equals vmax. The graph shown in the figure gives a plot of v2max varying with frequency n. v2max ν n l Obtain an expression for (a) Planck’s constant, and (b) The work function of the given photosensitive material in terms of the parameters ‘l’, ‘n’ and the mass ‘m’ of the electron. (c) How is threshold frequency determined from the plot? 3 Dual Nature of Matter and Radiation 483

12. Light of wavelength 2000 Å falls on a metal surface of work functions 4.2 eV. What is the kinetic energy (in eV) of the fastest electrons emitted from the surface? (i) What will be the change in the energy of the emitted electrons if the intensity of light with same wavelength is doubled? (ii) If the same light falls on another surface of work functions 6.5 eV, what will be the energy of emitted electrons?  3 13. Draw graphs showing the variation of photoelectric current with anode potential of a photocell for (i) same frequency but different intensities I1 > I2 > I3 of incident radiation. (ii) same intensity but different frequency υ1 > υ2 > υ3 of incident radiation. Explain why the saturation current is independent of the anode potential for incident radiation of different frequencies but same intensity. 5 Answers 1. (i) (d) (ii) (b) (iii) (a) 2. (i) intensity (ii) m = 12.27 A° V 7. (i) 5. 187 × 1014 Hz, (ii) 4536 Å zzz 484 Xam idea Physics–XII

Chapter –12 Atoms 1. Geiger-Marsden’s α-particle Scattering Experiment On the suggestion of Rutherford, in 1911, his two associates, H. Geiger and E. Marsden, performed an experiment by bombarding α-particles (Helium nuclei Z = 2, A = 4) on a gold foil. Observations: (i) Most of the α-particles pass through the gold foil undeflected. (ii) A very small number of α-particles (1 in 8000) suffered large angle deflection; some of them retraced their path or suffered 180° deflection. Conclusion: (i) Atom is hollow. (ii) Entire positive charge and nearly whole mass of atom is concentrated in a small centre called nucleus of Incident beam of atom. (iii) Coulomb’s law holds good α-particles Nucleus for atomic distances. (iv) Negatively charged electrons Detector are outside the nucleus. Impact Parameter: The perpendicular distance of initial Rutherford Scattering experiment velocity vector of α-particle from the nucleus, when the particle is far away from the nucleus, is called the impact parameter. It is denoted by b. For head on approach of α-particle, b = 0. Angle of Scattering (φ) : The angle by which α-particle is deviated from its original direction is called angle of scattering. EF b = 1 Ze2 cot z C 4rf0 EK 2 where Ek is the initial kinetic energy for head on approach of b' large alpha particle. A Bφ Impact parameter, b= 0. b G H +Ze N Atoms 485

2. Distance of Closest Approach The smallest distance of approach of α-particle near heavy nucleus is a measure of the size of nucleus. 2Ze2 Distance of nearest approach ≈ size of nucleus = 1 EK 4rf0 where EK is kinetic energy of incident α-particle, Z = atomic number, e = electronic charge. 3. Rutherford’s Atom Model Atom consists of a central heavy nucleus containing positive charge and negatively charged electrons circulating around the nucleus in circular orbits. Rutherford model could explain the neutrality of an atom, thermionic emission and photoelectric effect; but it could not explain the stability of an atom and the observed line spectrum of an atom (atomic spectrum). 4. Bohr’s Model Bohr modified Rutherford atom model to explain the line spectrum of hydrogen. Postulates of Bohr’s Theory (i) Stationary Circular Orbits: An atom consists of a central positively +r v charged nucleus and negatively charged electrons revolve around the nucleus in certain orbits called stationary orbits. + Ze m The electrostatic coulomb force between electrons and the nucleus provides the necessary centripetal force. i.e., mv2 = 1 (Ze) (e) …(i) r 4rf0 r2 where Z is the atomic number, m is the mass of electrons, r = radius of orbit. (ii) Quantum Condition: The stationary orbits are those in which angular momentum of electron h is an integral multiple of 2r , i.e., mvr = n h , n = 1, 2, 3,... …(ii) 2r Integer n is called the principal quantum number. This equation is called Bohr’s quantum condition. (iii) Transitions: The electron does not radiate energy when in a stationary orbit. The quantum of energy (or photon) is emitted or absorbed when an electron jumps from one stationary orbit to the other. The frequency of emitted or absorbed photon is given by hν = |Ei –Ef| …(iii) This is called Bohr’s frequency condition. Radius of Orbit and Energy of Electron in Orbit Condition of motion of electron in circular orbit is mv2 = 1 (Ze) (e) …(i) r 4rf0 r2 …(ii) Bohr’s quantum condition is mvr = n h 2r ⇒ v = nh 2rmr Substituting this value of v in (i), we get m c nh 2 = 1 Ze2 r 2rmr 4rf0 r2 m 486 Xam idea Physics–XII

This gives r = f0 h2 n2 rmZe2 Denoting radius of nth orbit by rn, we have f0 h2 n2 rn = rmZe2 …(iii) For hydrogen atom Z =1, ∴ (rn) H = f0h2 n2 rme2 The radius of first orbit of hydrogen atom is called Bohr’s radius. It is denoted by a0 ⇒ a0 = f0 h2 = 0.529 # 10–10 m = 0.529 Å rme2 Energy of Orbiting Electron From equation (i), mv2 = 1 Ze2 Kinetic energy, 4rf0 r Potential energy, K= 1 mv2 = 1 Ze2 2 4rf0 2r U = 1 (Ze) (– e) = – 1 Ze2 4rf0 r 4rf0 r Total energy E=K+U= 1 Ze2 – 1 Ze2 4rf0 2r 4rf0 r ⇒ E =– 1 Ze2 4rf0 2r For nth orbit, writing En for E, we have Ze2 1 2rn En =– 4rf0 …(iv) Substituting the value of rn from (iii) in (iv), we get mZ2 e4 Ze2 8f20 h2 n2 En =– 1 =– …(v) 4rf0 f0 h2 n2 …(vi) 2f rmZe2 p For convenience introducing Rydberg constant, R = me4 The value of Rydberg constant is 1.097 × 107 m–1. 8f02 ch3 We have En = – Z2 Rhc …(vii) n2 For hydrogen atom Z = 1, Energy of orbiting electron in H-atom En = – Rhc n2 ⇒ En = – 13.6 eV …(viii) n2 Equations (iii) and (vii) indicate that radii and energies of hydrogen like atoms (i.e., atoms containing one electron only) are quantised. Atoms 487

5. Energy Levels of Hydrogen Atom The energy of electron in hydrogen atom (Z = 1) is given (or series of hydrogen spectrum) by En = – Rhc = – 13.6 eV; n2 n2 Continuum E > 0 when n = 1, E1 = –13.6 eV n=∞ 0 eV when n = 2, n=7 –0.28 eV when n = 3, E2 = – 13.6 eV = – 3.4 eV n=6 –0.38 eV 4 n=5 –0.54 eV E3 = – 13.6 eV = –1.51 eV n=4 9 –0.85 eV when n = 4, E4 = – 13.6 eV = –0.85 eV n=3 –1.51 eV 16 when n = 5, E5 = – 13.6 eV = –0.54 eV n=2 –3.40 eV 25 when n = 6, E6 = – 13.6 eV = –0.38 eV 36 when n = 7, E7 = – 13.6 eV = –0.28 eV –13.60 eV 49 n=1 .................................................................. .................................................................. Fig. (a) Energy Level Diagram when n = ∞, E3 = – 13.6 eV = 0 eV (3) 2 If these energies are expressed by vertical lines on proper scale, the diagram obtained is called the energy level diagram. The energy level diagram of hydrogen atom is shown in fig. (a). Clearly the separation between lines goes on decreasing rapidly with increase of n (i.e., order of orbit). The series of lines of H-spectrum are shown in fig. (b). If the total energy of electron is above zero, the electron is free and can have any energy. Thus there is a continuum of energy states above E = 0 eV. 6. Hydrogen Spectrum Hydrogen emission spectrum consists of 5 series. (i) Lyman series: This lies in ultraviolet region. (ii) Balmer series: This lies in the visible region. (iii) Paschen series: This lies in near infrared region. (iv) Brackett series: This lies in mid infrared region. (v) Pfund series: This lies in far infrared region. Hydrogen absorption spectrum consists of only Lyman series. Explanation of Hydrogen Spectrum: ni and nf are the quantum numbers of initial and final states and Ei and Ef are energies of electron in H-atom (Z =1) in initial and final states then we have En erg y o f absorbedEpi h=ot–oRnnhi2c and Ef = – Rhc n 2 f TE = Ef – Ei = Rhc f 1 – 1 p n 2 n 2 f i If ν is the frequency of emitted radiation, we have from Bohr’s fourth postulate o= Ei –E f = – Rc –f– Rc p = Rcf 1 – 1 p …(ix) h n 2 n 2 n 2 n 2 i f f i 488 Xam idea Physics–XII

The wave number (i.e., reciprocal of wavelength) of the emitted radiation is given by o= 1 = o = Rf 1 – 1 m c ni2 p n 2 f The relation explains successfully the origin of various lines in the spectrum of hydrogen atom. The series of lines are obtained due to the transition of electron from various other orbits to a fixed inner orbit. Continuum n =∞ Pfund series 0 eV Brackett series –0.28 eV n=7 Paschen series –0.38 eV n=6 –0.54 eV n=5 –0.85 eV n=4 –1.51 eV n=3 n=2 Balmer series –3.40 eV Lyman series n=1 –13.6 eV Fig. (b) Series of H-spectrum (i) Lyman series: This series is produced when electron jumps from higher orbits to the first stationary orbit (i.e., nf =1). Thus for this series o= 1 = Re 1 – 1 o where ni = 2, 3, 4, 5,... m 12 n 2 i For longest wavelength of Lyman series ni = 2 1 1 1 3R ∴ mmax = Rd 12 – 22 n = 4 ∴ mmax = 4 = 4 m 3R 3 # 1.097 # 107    = 1.215×10–7 m = 1215 Å For shortest wavelength of Lyman series ni = ∞ 1 1 1 ∴ mmin = Rd 12 – 3 n = R mmin = 1 = 1 107 m = 0.9116 # 10 – 7 m = 911.6 Å R 1.097 # This is called series limit of Lyman series λlimit = 911.6 Å Obviously the lines of Lyman series are found in ultraviolet region. (ii) Balmer series: The series is produced when an electron jumps from higher orbits to the second stationary orbit (nf = 2). Thus for this series, o= 1 = Re 1 – 1 o where ni = 3, 4, 5, 6,... m 22 n 2 i For Longest wavelength of Balmer series (ni =3) 1 = Rd 1 – 1 n = 5R mmax 22 32 36 Atoms 489

mmax = 36 = 5 # 36 # 107 m = 6.563 # 10–7 m = 6563 Å 5R 1.097 For Shortest wavelength (or series limit) of Balmer series ni → ∞ 1 1 1 ∴ mmin = Rd 22 – 3 n = R 4 mmin = 4 = 4 10–7 m = 3.646 # 10–7 m =3646 Å R 1.097 # Obviously the lines of Balmer series are found in the visible region and first, second, third … lines are called Hα, Hβ, Hγ..., lines respectively. (iii) Paschen series: This series is produced when an electron jumps from higher orbits to the third stationary orbit (nf =3). 1 1 1 o = m = Re 32 – o where ni = 4, 5, 6, 7,... n 2 i For Longest wavelength of Paschen series (ni = 4) ∴ 1 = Rd 1 – 1 n = 7R mmax 32 42 144 ∴ mmax = 144 = 144 m = 18.752 # 10 –7 m =18752 Å 7R 7 # 1.097 # 107 For Series limit of Paschen series (ni =∞) 1 1 1 mmin = R d 32 – 3 n = R 9 mmin = 9 = 9 = 8.204 # 10–7 m= 8204 Å R 1.097 # 107 Obviously lines of Paschen series are found in infrared region. (iv) Brackett series: This series is produced when an electron jumps from higher orbits to the fourth stationary orbit (nf =4) 1 1 1 o = m = Re 42 – o where ni = 5, 6, 7, 8, ... n 2 i (v) Pfund series: This series is produced when an electron jumps from higher orbits to the fifth stationary orbit (nf = 5) o 1 1 1 = m = Re 52 – o where ni = 6, 7, 8, ... n 2 i The last three series are found in infrared region. The series spectrum of hydrogen atom is represented in figure. Selected NCERT Textbook Questions Q. 1. Suppose you are given a chance to repeat the alpha particle scattering experiment using a thin sheet of solid hydrogen in place of gold foil (hydrogen is a solid at temperature below 14 K). What results do you expect? Ans. Size of hydrogen nucleus = 1.2×10–15 m. ∴Electrostatic potential energy of a-particle at nuclear surface Ue = 1 (2e) (e) = 9 #109 # 2 # (1.6 # 10–19)2 J 4rf0 r 1.2 # 10–15 = 9 #109 # 2 #1.6 #10–19 eV 1.2 #10–15 = 2.4×106 eV = 2.4 MeV 490 Xam idea Physics–XII

This is much less than incident energy 5.5 MeV of α-particle; therefore α-particle will penetrate the nucleus and no scattering will be observed. Aliter: The de Broglie wavelength of α-particle is much less than inter-proton distance in solid hydrogen, so α-particle will move directly penetrating the nucleus. Q. 2. A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes transition from the upper level to the lower level? Ans. According to Bohr’s postulate E1 – E2 = hν ∴ Frequency of emitted radiation ν = E1 − E2 = 2.3 eV = 5.55×1014 Hz = 2.3 #h 1.6 # 10h–19 J 6.63 # 10–34 J - s Q. 3. The ground state energy of hydrogen atom is –13.6 eV. What is the kinetic and potential energies of the electron in the ground and second excited state? [CBSE (AI) 2010, 2011, Bhubaneshwar 2015] Ans. Kinetic energy, K = 1 mv2 = 1 . e2 [for H-atom, Z = 1] …(i) 2 4rf0 2r Potential energy, U = – 1 e2 …(ii) 4rf0 r Total energy   E = K +U = − 1 e2 …(iii) 4πε0 2r Comparing equations (i), (ii), (iii), we have K = – E and U = 2E Given      E = –13.6 eV (For ground state n = 1) ∴ Kinetic energy, K = 13.6 eV Potential energy U = 2×(–13.6 eV)=–27.2 eV For second excited state, n=3 ∴ K= −E = +13.6 eV = 1.51 eV 9 and      U = 2E = 2 # (–13.6 eV) = – 3.02 eV 9 Q. 4. A hydrogen atom initially in the ground state absorbs a photon, which excites it to the n=4 level. Determine the wavelength and frequency of photon. Ans. The energy levels of H-atom are given by En = − Rhc n2 For given transition n1=1, n2=4 ∴ E1 = − Rhc , E2 = − Rhc 12 42 ∴ Energy of absorbed photon ∆E = E2 − E1 = Rhc  1 − 1  or  12 42  ∆E = 15 Rhc …(i) 16 Atoms 491

∴ Wavelength of absorbed photon λ is given by ∆E = hc λ ∴ hc = 15 Rhc ⇒ λ = 16 or λ 16 15R 16 m = 15 # 1.097 # 107 m = 9.72 # 10–8 m Frequency, o = c = 3 #108 = 3.09 # 1015 Hz m 9.72 #10–8 Q. 5. (a) Using the Bohr’s model, calculate the speed of electron in the hydrogen atom in n=1, 2 and 3 levels. (b) Calculate the orbital period in each of these levels. Ans. (a) The speed of electron in stable orbit of H-atom is v= e2 h . 1 = (1.6 ×10−19 )2 1 2ε0 n 2 × 8.85 ×10−12 × 6.63×10−34  n  = 2.18 ×106 m /s n For n=1, v1 = 2.18 × 106 m/s. For n=2, v2 = 2.18 # 106 = 1.09 # 106 m/s 2 For n=3, v3 = 2.18 # 106 = 7.27 # 105 m/s 3 Obviously the speed of electron goes on decreasing with increasing n. (b) Time period, T = 2πr = 2π(ε0 h2n2 / πme2 ) v (e2 / 2ε0 hn)   = 4ε02 h3n3 = 4 × (8.85 ×10−12 )2 × (6.63×10−34 )3 × n3 me4 9.1 ×10−31 × (1.6 ×10−19 )4   = 1.53×10–16 n3 seconds For n=1, T1 = 1.53×10–16 s For n=2, T2 = 1.53×10–16×(2)3 = 12.24×10–16 s For n=3, T3 = 1.53×10–16×(3)3 = 41.31×10–16 s Q. 6. The radius of innermost orbit of a hydrogen atom is 5.3×10–11 m. What are the radii of n=2 and n=3 orbits? Ans. The radii of Bohr’s orbits are given by   rn = fr0 hm2en22 & rn ? n2 For ground state n = 1, r1 = 5.3 × 10–11 m (given)   rr12 = d n2 2 n1 n ⇒    r2 = c 2 2 = 4r1 = 4 # 5.3 # 10–11 = 2.12 # 10–10 m 1 m r1 For n =3,  r3 = (3)2 r1 = 9 × 5.3 × 10–11    = 4.77 × 10–10 m 492 Xam idea Physics–XII

Q. 7. In accordance with 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) Ans. According to Bohr’s model, angular momentum 2rmvr h h mvr = n 2r ⇒ n= Given m = 6.0 × 1024 kg, v = 3 × 104 m/s, r = 1.5 × 1011 m ∴ n = 2×3.14 # 6 #1024 ×3×104 ×1.5×1011 = 2.57 ×1074 6.6 ×10–34 Q. 8. 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 207 me orbits around a proton]. Ans. If mµ is the mass of muon, then from Bohr’s theory 1 e2 = mn v2 and mn vr = nh [for H-atom, Z = 1] 4rf0 r2 r 2r Eliminating v from these equations, we get r = ε0 h2n2 πmµ e2 As mµ =207me,where me is mass of electron ∴ r = ε0 h2n2 207πme e2 For ground state for muon, we have rn = f0 h2 207rme .e2 But f0 h2 = ground state radius of H-atom = 0.53×10–10 m rme e2 ∴ rn = 0.53 # 10–10 = 2.56 # 10–13 m 207 me4 1 Also energy     En = − 8ε02 h2 . n2 Obviously, En ? m En = mn & En = mn # Ee Ee me me Ground state energy of an electron in H-atom, Ee = – 13.6 eV ∴ En = 207me ×(–13.6 eV) = – 2.8 × 103 eV = – 2.8 keV me Multiple Choice Questions [1 mark] Choose and write the correct option(s) in the following questions. 1. The size of the atom is proportional to (a) A (b) A1/3 (c) A2/3 (d) A–1/3 2. To explain his theory, Bohr used (b) quantisation of angular momentum (a) conservation of linear momentum (d) none of these (c) conservation of quantum Atoms 493

3. Taking the Bohr radius as a0 = 53 pm, the radius of Li++ ion in its ground state, on the basis of Bohr’s model, will be about [NCERT Exemplar] (a) 53 pm (b) 27 pm (c) 18 pm (d) 13 pm 4. The ratio of energies of the hydrogen atom in its first to second excited state is 1 1 (a) 1 : 4 (b) 4 : 1 (c) – 4 : – 9 (d) – 4 : – 9 5. The binding energy of a H-atom, considering an electron moving around a fixed nuclei me4 (proton), is B = – 8n2 f02 h2 (m = electron mass). If one decides to work in a frame of reference where the electron is at rest, the proton would be moving arround it. By similar arguments, the binding energy would be B = – me4 (M = proton mass) [NCERT Exemplar] 8n2 f02 h2 This last expression is not correct because (a) n would not be integral (b) Bohr-quantisation applies only to electron (c) the frame in which the electron is at rest is not inertial (d) the motion of the proton would not be in circular orbits, even approximately 6. The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons. This is because [NCERT Exemplar] (a) of the electrons not being subject to a central force (b) of the electrons colliding with each other (c) of screening effects (d) the force between the nucleus and an electron will no longer be given by Coulomb’s law 7. The ratio of the speed of the electrons in the ground state of hydrogen to the speed of light in vacuum is (a) 1/2 (b) 2/237 (c) 1/137 (d) 1/237 8. For the ground state, the electron in the H-atom has an angular momentum = h, according to the simple Bohr model. Angular momentum is a vector and hence there will be infinitely many orbits with the vector pointing in all possible directions. In actuality, this is not true, [NCERT Exemplar] (a) because Bohr model gives incorrect values of angular momentum. (b) because only one of these would have a minimum energy. (c) angular momentum must be in the direction of spin of electron. (d) because electrons go around only in horizontal orbits. 9. O2 molecule consists of two oxygen atoms. In the molecule, nuclear force between the nuclei of the two atoms [NCERT Exemplar] (a) is not important because nuclear forces are short-ranged. (b) is as important as electrostatic force for binding the two atoms. (c) cancels the repulsive electrostatic force between the nuclei. (d) is not important because oxygen nucleus have equal number of neutrons and protons. 10. In the following transitions of the hydrogen atom, the one which gives an absorption line of highest frequency is (a) n = 1 to n = 2 (b) n = 3 to n = 8 (c) n = 2 to n = 1 (d) n = 8 to n = 3 11. The wavelength of the first line of Lyman series in hydrogen is 1216 Å. The wavelength of the second line of the same series will be (a) 912 Å (b) 1026 Å (c) 3648 Å (d) 6566 Å 494 Xam idea Physics–XII

12. Two H atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced is [NCERT Exemplar] (a) 10.20 eV (b) 20.40 eV (c) 13.6 eV (d) 27.2 eV 13. When an electron in an atom goes from a lower to a higher orbit, its (a) kinetic energy (KE) increases, potential energy (PE) decreases (b) KE increases, PE increases (c) KE decreases, PE increases (d) KE decreases, PE decreases 14. According to Bohr’s theory, the energy of radiation in the transition from the third excited state to the first excited state for a hydrogen atom is (a) 0.85 eV (b) 13.6 eV (c) 2.55 eV (d) 3.4 eV 15. Given the value of Rydberg constant is 107 m–1, the wave number of the last line of the Balmer series in hydrogen spectrum will be (a) 0.25 × 107 m–1 (b) 2.5 × 107 m–1 (c) 0.025 × 104 m–1 (d) 0.5 × 107 m–1 16. If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength l. When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be 9 20 20 16 16 7 13 (a) 25 m (b) m (c) m (d) m 17. Hydrogen H, deuterium D, singly-ionised helium He+ and doubly-ionised lithium Li++ all have one electron around the nucleus. Consider n = 2 to n = 1 transition. The wavelengths of the emitted radiations are m1, m2, m3, and m4 respectively. Then approximately (a) m1 = 2m2 = 2 2 m3 = 3 2 m4 (b) m1 = m2 = 2m3 = 3m4 (c) m1 = m2 = 4m3 = 9m4 (d) 4m1 = 2m2 = 2m3 = m4 18. The Bohr model for the spectra of a H-atom [NCERT Exemplar] (a) will not be applicable to hydrogen in the molecular from. (b) will not be applicable as it is for a He-atom. (c) is valid only at room temperature. (d) predicts continuous as well as discrete spectral lines. 19. Let En = 1 me4 be the energy of the nth level of H-atom. If all the H-atoms are in the 8f02 n2 h2 ground state and radiation of frequency (E2 – E1)/h falls on it, [NCERT Exemplar] (a) it will not be absorbed at all. (b) some of atoms will move to the first excited state. (c) all atoms will be excited to the n = 2 state. (d) no atoms will make a transition to the n = 3 state. 20. A set of atoms in an excited state decays. (a) in general to any of the states with lower energy. (b) into a lower state only when excited by an external electric field. (c) all together simultaneously into a lower state. (d) to emit photons only when they collide. Answers 1. (b) 2. (b) 3. (c) 4. (d) 5. (c) 6. (a) 7. (c) 8. (a) 9. (a) 10. (a) 11. (b) 12. (a) 13. (c) 14. (c) 15. (a) 16. (c) 17. (c) 18. (a), (b) 19. (b), (d) 20. (a) Atoms 495

Fill in the Blanks [1 mark] 1. The angle of scattering θ for zero value of impact parameter b is _________________. 2. The frequency spectrum of radiation emitted as per Rutherford’s model of atom is _____________. 3. The force responsible for scattering of alpha particle with target nucleus is _______________. 4. According to de Broglie a stationary orbit is that which contains an _______________ number of de Broglie waves associated with the revolting electron. 5. _______________ is a physical quantity whose dimensions are the same as that of Plank’s constant. 6. _______________ series of hydrogen spectrum lies in the visible region electromagnetic spectrum. 7. _______________ is the ionisation potential of hydrogen atom. 8. Total energy of electron in a stationary orbit is _________________, which means the electron is bound to the nucleus and is not free to leave it. 9. The value of Rydberg constant is _________________. 10. When an electron jumps from 2nd stationary orbit of hydrogen atom to 1st stationary orbit, the energy emitted is _________________. Answers 1. 180° 2. continuous 3. electrostatic force 4. integral 7. 13.6 eV 5. Angular momentum 6. Balmer 10. 10.2 eV 8. negative 9. 1.09 × 107 m–1 Very Short Answer Questions [1 mark] Q. 1. Write the expression for Bohr’s radius in hydrogen atom. [CBSE Delhi 2010] Ans. Bohr’s radius, r1 = f0 h2 = 0.529×10–10 m rme2 Q. 2. In the Rutherford scattering experiment the distance of closest approach for an α-particle is d0. If α-particle is replaced by a proton, how much kinetic energy in comparison to α-particle will it require to have the same distance of closest approach d0? [CBSE (F) 2009] Ans. Ek = 1 ]Zeg]2eg (for α-particle, q = 2e) 4rf0 d0 Ekl = 1 ]Zeg] e g (for proton, q = e) 4rf0 d0 EEklk = Ek 1 ⇒ Ekl = 2 2 That is KE of proton must be half on comparison with KE of α-particle. Q. 3. What is the ratio of radii of the orbits corresponding to first excited state and ground state in a hydrogen atom? [CBSE Delhi 2010] Ans. rn = f0 h2 n2 \\ n2 rme2 For 1st excited state, n = 2 For ground state, n = 1 r2 ∴ r1 = 4 1 496 Xam idea Physics–XII

Q. 4. Find the ratio of energies of photons produced due to transition of an electron of hydrogen atom from its: (i) second permitted energy level to the first level, and (ii) the highest permitted energy level to the first permitted level. [CBSE (AI) 2010] Ans. EI = Rhce 1 – 1 o = 3 Rhc 12 22 4 EII = Rhcd 1 – 1 n = Rhc 12 32 Ratio EI = 3 EII 4 Q. 5. State Bohr’s quantisation condition for defining stationary orbits. [CBSE (F) 2010] Ans. Quantum Condition: The stationary orbits are those in which angular momentum of electron is h an integral multiple of 2r i.e., mvr = n h , n = 1, 2, 3, ... 2r Integer n is called the principal quantum number. This equation is called Bohr’s quantum condition. Q. 6. The radius of innermost electron orbit of a hydrogen atom is 5.1 × 10–11 m. What is the radius of orbit in the second excited state? [CBSE Delhi 2010] Ans. In ground state, n = 1 In second excited state, n = 3 As rn ∝ n2 ∴ r3 r1 = c 3 2 = 9 1 m r3 = 9r1 = 9 × 5.1 × 10–11 m = 4.59 × 10–10 m Q. 7. The mass of H-atom is less than the sum of the masses of a proton and electron. Why is this so? [NCERT Exemplar] [HOTS] Ans. Einstein’s mass-energy equivalence gives E = mc2. Thus the mass of an H-atom is mp+me – B C2 where B ≈ 13.6 eV is the binding energy. It is less than the sum of masses of a proton and an electron. Q. 8. When an electron falls from a higher energy to a lower energy level, the difference in the energies appears in the form of electromagnetic radiation. Why cannot it be emitted as other forms of energy? [NCERT Exemplar] [HOTS] Ans. This is because electrons interact only electromagnetically. Q. 9. Would the Bohr formula for the H-atom remain unchanged if proton had a charge (+4/3)e and electron had a charge (–3/4)e, where e = 1.6 × 10–19 C? Give reasons for your answer. [NCERT Exemplar] [HOTS] Ans. Yes, since the Bohr formula involves only the product of the charges. Q. 10. Consider two different hydrogen atoms. The electron in each atom is in an excited state. Is it possible for the electrons to have different energies but the same orbital angular momentum according to the Bohr model? [NCERT Exemplar] [HOTS] 13.6 Ans. No, because according to Bohr model, En = – n2 , and electrons having different energies belong to different levels having different values of n. So, their angular momenta will be different, as mvr = nh . 2r Atoms 497