DC Pandey Optics And Modern Physics

Solved Examples TYPED PROBLEMS Type 1. Based on de-Broglie wavelength V Example 1 A proton is fired from very far away towards a nucleus with charge Q = 120 e, where e is the electronic charge. It makes a closest approach of 10 fm to the nucleus. The de-Broglie wavelength (in units of fm) of the proton at its start is [Take the proton mass, mp = (5/3) ´ 10-27 kg, h/e = 4.2 ´ 10-15 J-s/C, 1 = 9 ´ 109 m/F, 1 fm = 10-15 m] (JEE 2013) 4pe 0 Solution r = closest distance = 10 fm Ze e Å From energy conservation, we have + K Ki + Ui = Kf + U f r or K + 0 =0 + 1 . q1q2 4pe0 r or K = 1 . (120 e) (e) …(i) 4pe0 r de-Broglie wavelength, l= h …(ii) 2Km Substituting the given values in above two equations, we get l = 7 ´ 10-15 m = 7 fm V Example 2 Find de-Broglie wavelength of single electron in 2 nd orbit of hydrogen atom by two methods. Solution Method 1 Kinetic energy of single electron in 2nd orbit is 3.4 eV using the equation, l = 150 = 150 = 6.64 Å Ans. KE (in eV ) 3.4 Method 2 Circumference of nth orbit = nl \\ 2 pr = 2 l or l = pr Now, r µ n2 r1 = 0.529 Å \\ l = p(0.529)(2)2Å = 6.64 Å Ans.

Chapter 33 Modern Physics - I — 289 Type 2. Based on Bohr’s atomic models V Example 3 The electric potential between a proton and an electron is given by V = V0 ln r , where r0 is a constant. Assuming Bohr model to be applicable, r0 write variation of rn with n, being the principal quantum number. (JEE 2003) (a) rn µ n (b) rn µ 1 (c) rn µ n 2 (d) rn µ 1 n n2 Solution Q U = eV = eV0 ln èæçç r öø÷÷ r0 |F |= - dU = eV0 dr r This force will provide the necessary centripetal force. Hence, mv2 = eV0 rr or v = eV0 …(i) m Moreover, mvr = nh …(ii) 2p Dividing Eq. (ii) by Eq. (i), we have mr = èçæ nh öø÷ m or rn µ n 2p eV0 V Example 4 Imagine an atom made up of proton and a hypothetical particle of double the mass of the electron but having the same charge as the electron. Apply the Bohr atom model and consider all possible transitions of this hypothetical particle to the first excited level. The longest wavelength photon that will be emitted has wavelength l (given in terms of the Rydberg constant R for the hydrogen atom) equal to (JEE 2000) (a) 9/5R (b) 36/5R (c) 18/5R (d) 4/R Solution In hydrogen atom, En = - Rhc n2 Also, En µm where, m is the mass of the electron. Here, the electron has been replaced by a particle whose mass is double of an electron. Therefore, for this hypothetical atom energy in nth orbit will be given by En = - 2Rhc n2 The longest wavelength lmax (or minimum energy) photon will correspond to the transition of particle from n = 3 to n = 2 . \\ hc = E3 - E 2 = 2Rhc èæç 1 - 1 ÷øö lmax 22 32 This gives, lmax = 18 /5R \\ The correct option is (c).

290 — Optics and Modern Physics V Example 5 The recoil speed of a hydrogen atom after it emits a photon is going from n = 5 state to n = 1 state is ........ m/s. (JEE 1997) Solution From conservation of linear momentum, |Momentum of recoil hydrogen atom| = |Momentum of emitted photon| or mv = DE c Here, DE = E5 - E1 = – 13.6 é 1 - 1 ù eV ëê 52 12 úû = (13.6) (24 /25) eV = 13.056 eV = 13.056 ´ 1.6 ´ 10-19 J = 2.09 ´ 10-18 J and m = mass of hydrogen atom = 1.67 ´ 10-27 kg \\ v = DE = 2.09 ´ 10-18 mc (1.67 ´ 10-27 ) (3 ´ 108 ) v » 4.17 m/s V Example 6 A hydrogen like atom (described by the Bohr model) is observed to emit six wavelengths, originating from all possible transitions between a group of levels. These levels have energies between – 0.85 eV and –0.544 eV (including both these values). (JEE 2002) (a) Find the atomic number of the atom. (b) Calculate the smallest wavelength emitted in these transitions. (Take, hc = 1240 eV-nm, ground state energy of hydrogen atom = -13.6 eV) Solution (a) Total 6 lines are emitted. Therefore, n =4 n (n - 1) = 6 or 2 So, transition is taking place between mth energy state and (m + 3) th energy state. Em = - 0.85 eV or -13.6 çèæç Z2 øö÷÷ = - 0.85 m2 or Z = 0.25 …(i) m Similarly, Em + 3 = - 0.544 eV or -13.6 Z2 = - 0.544 (m + 3)2 or Z = 0.2 …(ii) (m + 3) Solving Eqs. (i) and (ii) for Z and m, we get m = 12 and Z = 3 Ans. (b) Smallest wavelength corresponds to maximum difference of energies which is obviously Em + 3 - Em \\ D Emax = - 0.544 - (- 0.85) = 0.306 eV \\ lmin = hc = 1240 = 4052.3 nm Ans. DE max 0.306

Chapter 33 Modern Physics - I — 291 V Example 7 A hydrogen like atom of atomic number Z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition to quantum state n, a photon of energy 40.8 eV is emitted. Find n, Z and the ground state energy (in eV) of this atom. Also calculate the minimum energy (in eV) that can be emitted by this atom during de-excitation. Ground state energy of hydrogen atom is –13.6 eV. (JEE 2000) Solution Let ground state energy (in eV) be E1. Then, from the given condition E2n - E1 = 204 eV or E1 - E1 = 204 eV 4n2 or E1 èæç 1 - 1øö÷ = 204 eV …(i) 4n2 and E2n - En = 40.8 eV or E1 - E1 = 40.8 eV 4n2 n2 or E1 æèç -3 öø÷ = 40.8 eV …(ii) 4n2 From Eqs. (i) and (ii), we get 1 - 1 2 1 15 or 4n 4n2 4n2 =5 or 1 = + 3 4n2 4 = 1 or n =2 n2 From Eq. (ii), E1 = - 4 n2 (40.8) eV 3 = - 4 (2)2 (40.8) eV 3 or E1 = - 217.6 eV E1 = - (13.6) Z 2 \\ Z2 = E1 = -217.6 = 16 -13.6 -13.6 \\ Z =4 E min = E2n - E 2n -1 = E1 - E1 4n2 (2n - 1)2 = E1 - E1 = - 7 E1 16 9 144 = - æçè 7 öø÷ (- 217.6) eV 144 \\ Emin = 10.58 eV

292 — Optics and Modern Physics V Example 8 A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n. The excited atom can make a transition to the first excited state by successively emitting two photons of energy 10.2 eV and 17.0 eV, respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies 4.25 eV and 5.95 eV, respectively. (JEE 1994) Determine the values of n and Z. (Ionization energy of H-atom = 13.6 eV) Solution From the given conditions, and En - E2 = (10.2 + 17) eV = 27.2 eV …(i) Eq. (i) - Eq. (ii) gives En - E3 = (4.25 + 5.95) eV = 10.2 eV …(ii) Þ E3 - E2 = 17.0 eV or Z 2 (13.6) èæç 1 - 1 øö÷ = 17.0 Þ 4 9 Z 2 (13.6) (5 /36) = 17.0 Z 2 = 9 or Z = 3 From Eq. (i), Z2 (13.6) æçè 1 - 1 øö÷ = 27.2 or (3)2 (13.6) æèç 1 - 1 öø÷ = 27.2 4 n2 4 n2 or or 1 - 1 = 0.222 or 1 / n2 = 0.0278 4 n2 n2 = 36 Þ n = 6 Type 3. Based on X-rays V Example 9 Determine the energy of the characteristic X-ray ( Kb ) emitted from a tungsten ( Z = 74) target when an electron drops from the M-shell ( n = 3) to a vacancy in the K-shell ( n = 1). Solution Energy associated with the electron in the K-shell is approximately EK = – (74 – 1)2 (13.6 eV ) = – 72474 eV An electron in the M-shell is subjected to an effective nuclear charge that depends on the number of electrons in the n = 1 and n = 2 states because these electrons shield the M electrons from the nucleus. Because there are eight electrons in the n = 2 state and one remaining in the n = 1 state, roughly nine electrons shield M electrons from the nucleus. So, Z eff = z – 9 Hence, the energy associated with an electron in the M-shell is EM = –13.6 Z 2 eV = –13.6 (Z – 9)2 eV eff 32 32 = – (13.6) (74 – 9)2 eV = – 6384 eV 9 Therefore, emitted X-ray has an energy equal to Ans. EM – EK = {–6384 – (–72474)} eV = 66090 eV

Chapter 33 Modern Physics - I — 293 V Example 10 The potential difference applied to an X-ray tube is 5 kV and the current through it is 3.2 mA. Then, the number of electrons striking the target per second is (JEE 2002) (a) 2 ´ 1016 (b) 5 ´ 106 (c) 1 ´ 1017 (d) 4 ´ 1015 Solution Q i = q = ne n=it tt e \\ i = 3.2 ´10-3 A, Substituting e = 1.6 ´10-19 C and t = 1 s We get, n = 2 ´ 1016 \\ The correct answer is (a). V Example 11 X-rays are incident on a target metal atom having 30 neutrons. The ratio of atomic radius of the target atom and 4 He is (14)1/3 . (JEE 2005) 2 (a) Find the mass number of target atom. (b) Find the frequency of K a line emitted by this metal. Hint : Radius of a nucleus (r) has the following relation with mass number (A). r µ A1/3 (R = 1.1 ´ 107 m-1, c = 3 ´ 108 m/s) Solution (a) From the relation r µ A1/3 , We have, r2 = æçèç A2 ÷÷øö1/3 r1 A1 or æç A2 ö÷1/3 = (14)1/3 è4ø \\ A2 = 56 (b) Z 2 = A2 - number of neutrons = 56 - 30 = 26 \\ f = Rc (Z - 1)2 çèæ 1 - 1 ø÷ö = 3Rc (Z - 1)2 12 22 4 Substituting the given values of R, c and Z, we get f = 1.55 ´ 1018 Hz V Example 12 Stopping potential of 24, 100, 110 and 115 kV are measured for photoelectrons emitted from a certain element when it is radiated with monochromatic X-ray. If this element is used as a target in an X-ray tube, what will be the wavelength of K a -line? Solution Stopping potentials are 24, 100, 110 and 115 kV, i.e. if the electrons are emitted from conduction band, maximum kinetic energy of photoelectrons would be 115 ´ 103 eV. If they are emitted from next inner shell, maximum kinetic energy of photoelectrons would be 110 ´ 103 eV and so on.

294 — Optics and Modern Physics For photoelectrons of L- shell it would be 100 ´ 103 eV and for K-shell it is 24 ´ 103 eV. Therefore, difference between energy of L-shell and K-shell is DE = EL – EK = (100 – 24) ´ 103 eV = 76 ´ 103 eV \\ Wavelength of K a-line (transition of electron from L-shell to K-shell) is, lK a (in Å ) = 12375 ) = 12375 DE (in eV 76 ´ 103 = 0.163 Å Ans. V Example 13 In Moseley’s equation f = a ( Z – b), a and b are constants. Find their values with the help of the following data. Element Z Wavelength of Ka X-rays Mo 42 0.71 Å Co 27 1.785 Å Solution f = a (Z – b) or c =a (Z1 – b) …(i) l1 …(ii) …(iii) and c = a (Z 2 – b) l2 Ans. From Eqs. (i) and (ii), we have é 1– 1 ù = a (Z1 – Z 2) cê l1 l2 ú û ë Solving above three equations with c = 3.0 ´ 108 m/s, l1 = 0.71 ´ 10–10 m l2 = 1.785 ´ 10–10 m, Z1 = 42 and Z 2 = 27, we get a = 5 ´ 107 (Hz)1/2 and b = 1.37 Type 4. Based on photoelectric effect V Example 14 A monochromatic light source of frequency f illuminates a metallic surface and ejects photoelectrons. The photoelectrons having maximum energy are just able to ionize the hydrogen atoms in ground state. When the whole experiment is repeated with an incident radiation of frequency 5 f, the 6 photoelectrons so emitted are able to excite the hydrogen atom beam which then emits a radiation of wavelength 1215 Å. (a) What is the frequency of radiation? (b) Find the work-function of the metal. Solution (a) Using Einstein’s equation of photoelectric effect, K max = hf – W

Chapter 33 Modern Physics - I — 295 Here, K max = 13.6 eV …(i) \\ hf – W = 13.6 eV …(ii) Further, h èçæ 5 f ø÷ö –W = 12375 = 10.2 eV Ans. 6 1215 [from Eq. (i)] Solving Eqs. (i) and (ii), we have Ans. hf = 3.4 eV or f = (6) (3.4) (1.6 ´ 10–19 ) 6 (6.63 ´ 10–34 ) = 4.92 ´ 1015 Hz (b) W = hf – 13.6 = 6 (3.4) – 13.6 = 6.8 eV V Example 15 The graph between 1/l and stopping potential (V ) of three metals having work-functions f 1 , f 2 and f 3 in an experiment of photoelectric effect is plotted as shown in the figure. Which of the following statement(s) is/are correct? (Here, l is the wavelength of the incident ray). (JEE 2006) V Metal 1 Metal 2 Metal 3 q 0.001 0.002 0.004 1/l nm–1 (a) Ratio of work-functions f1 : f2 : f3 = 1 : 2 : 4 (b) Ratio of work-functions f1 : f2 : f3 = 4 : 2 : 1 (c) tan q is directly proportional to hc / e, where h is Planck constant and c is the speed of light (d) The violet colour light can eject photoelectrons from metals 2 and 3 Solution From the relation, eV = hc - f or V = çæè hec÷öø æèç 1 ÷øö - f l l e This is equation of straight line. Slope is tan q = hc . e Further V = 0 at f = hc l \\ f1 : f2 : f3 = hc : hc : hc = 1 : 1 : 1 =1:2:4 l01 l02 l03 l01 l02 l03 1 = 0.001 nm-1 or l01 = 10000 Å l01 1 = 0.002 nm-1 or l02 = 5000 Å l02 1 = 0.004 nm-1 or l03 = 2500 Å l03 Violet colour has wavelength 4000 Å. So, violet colour can eject photoelectrons from metal 1 and metal 2. \\ The correct options are (a) and (c).

296 — Optics and Modern Physics V Example 16 A beam of light has three wavelengths 4144 Å, 4972 Å and 6216 Å with a total intensity of 3.6 ´ 10-3 Wm -2 equally distributed amongst the three wavelengths. The beam falls normally on an area 1.0 cm2 of a clean metallic surface of work-function 2.3 eV. Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated in two seconds. (JEE 1989) Solution Energy of photon having wavelength 4144 Å, E1 = 12375 eV 4144 = 2.99 eV Similarly, E2 = 12375 eV 4972 = 2.49 eV and E3 = 12375 eV 6216 = 1.99 eV Since, only E1 and E2 are greater than the work-function W = 2.3 eV , only first two wavelengths are capable for ejecting photoelectrons. Given intensity is equally distributed in all wavelengths. Therefore, intensity corresponding to each wavelength is 3.6 ´ 10-3 = 1.2 ´ 10-3 W/m2 3 Or energy incident per second in the given area (A = 1.0 cm2 = 10-4 m2) is P = 1.2 ´ 10-3 ´ 10-4 = 1.2 ´ 10-7 J/s Let n1 be the number of photons incident per unit time in the given area corresponding to first wavelength. Then, n1 = P E1 = 1.2 ´ 10-7 2.99 ´ 1.6 ´ 10-19 = 2.5 ´ 1011 Similarly, n2 = P E2 = 1.2 ´ 10-7 2.49 ´ 1.6 ´ 10-19 = 3.0 ´ 1011 Since, each energetically capable photon ejects one electron, total number of photoelectrons liberated in 2 s. = 2(n1 + n2) = 2 (2.5 + 3.0) ´ 1011 = 1.1 ´ 1012 Ans.

Miscellaneous Examples V Example 17 Two metallic plates A and B each of area 5 ´ 10-4 m2 , are placed parallel to each other at a separation of 1 cm. Plate B carries a positive charge of 33.7 ´ 10-12 C. A monochromatic beam of light, with photons of energy 5 eV each, starts falling on plate A at t = 0 so that 1016 photons fall on it per square metre per second. Assume that one photoelectron is emitted for every 106 incident photons. Also assume that all the emitted photoelectrons are collected by plate B and the work-function of plate A remains constant at the value 2 eV. Determine (JEE 2002) (a) the number of photoelectrons emitted upto t = 10 s, (b) the magnitude of the electric field between the plates A and B at t = 10 s and (c) the kinetic energy of the most energetic photoelectrons emitted at t = 10s when it reaches plate B. Neglect the time taken by the photoelectron to reach plate B. (Take, e0 = 8.85 ´ 10-12C2/ N -m2). Solution Area of plates, = 5 ´ 10-4 m2 Distance between the plates, d = 1 cm = 10-2 m (a) Number of photoelectrons emitted upto t = 10 s are n = (number of photons falling on unit area in unit time) ´ (area ´ time) 106 = 1 [(10)16 ´ (5 ´ 10-4 ) ´ (10)] 106 = 5.0 ´ 107 Ans. (b) At time t = 10 s, qA = + ne = (5.0 ´ 107 ) (1.6 ´ 10-19 ) Charge on plate A, = 8.0 ´ 10-12 C and charge on plate B, qB = (33.7 ´ 10-12 - 8.0 ´ 10-12) = 25.7 ´ 10-12 C \\ Electric field between the plates, E = (qB - qA) 2 Ae0 or E = 2 ´ (25.7 - 8.0) ´ 10-12 (5 ´ 10-4 ) (8.85 ´ 10-12) = 2 ´ 103 N/C (c) Energy of most energetic photoelectrons at plate A, = E - W = (5 - 2) eV = 3 eV

298 — Optics and Modern Physics Increase in energy of photoelectrons Ans. = (eEd) joule = (Ed) eV = (2 ´ 103 ) (10-2) eV = 20 eV Energy of photoelectrons at plate B = (20 + 3) eV = 23 eV V Example 18 Photoelectrons are emitted when 400 nm radiation is incident on a surface of work-function 1.9 eV. These photoelectrons pass through a region containing a-particles. A maximum energy electron combines with an a-particle to form a He + ion, emitting a single photon in this process. He+ ions thus formed are in their fourth excited state. Find the energies in eV of the photons lying in the 2 to 4 eV range, that are likely to be emitted during and after the combination. [Take, h = 4.14 ´ 10-15 eV-s] (JEE 1999) Solution Given work-function W = 1.9 eV Wavelength of incident light, l = 400 nm \\ Energy of incident light, E = hc = 3.1 eV l (Substituting the values of h, c and l) Therefore, maximum kinetic energy of photoelectrons K max = E - W = (3.1 - 1.9) = 1.2 eV Now, the situation is as shown in figure. e– Kmax = 1.2eV n=5 E5 = – 2.2 eV a-particles He+ in fourth He+ excited state or n=5 (Z = 2) Energy of electron in 4th excited state of He+ (n = 5) will be E5 = - 13.6 Z2 eV n2 Þ E5 = - (13.6) (2)2 = - 2.2 eV (5)2 Therefore, energy released during the combination = 1.2 - (-2.2) = 3.4 eV Similarly, energies in other energy states of He+ will be E4 = - 13.6 (2)2 = - 3.4 eV (4)2 E3 = - 13.6 (2)2 = - 6.04 eV (3)2 E2 = - 13.6 (2)2 = - 13.6 eV (2)2

Chapter 33 Modern Physics - I — 299 The possible transitions are DE5 ® 4 = E5 - E4 = 1.2 eV < 2 eV DE5 ®3 = E5 - E3 = 3.84 eV DE5 ® 2 = E5 - E2 = 11.4 eV > 4 eV DE4 ®3 = E4 - E3 = 2.64 eV DE4 ® 2 = E4 - E2 = 10.2 eV > 4 eV Hence, the energy of emitted photons in the range of 2 eV and 4 eV are 3.4 eV during combination and 3.84 eV and 2.64 after combination. V Example 19 Light from a discharge tube containing hydrogen atoms falls on the surface of a piece of sodium. The kinetic energy of the fastest photoelectrons emitted from sodium is 0.73 eV. The work-function for sodium is 1.82 eV. (JEE 1992) Find (a) the energy of the photons causing the photoelectrons emission. (b) the quantum numbers of the two levels involved in the emission of these photons. (c) the change in the angular momentum of the electron in the hydrogen atom, in the above transition, and (d) the recoil speed of the emitting atom assuming it to be at rest before the transition. (Ionization potential of hydrogen is 13.6 eV.) Solution (a) From Einstein’s equation of photoelectric effect, Energy of photons causing the photoelectric emission Ans. = Maximum kinetic energy of emitted photons + work-function or E = K max + W = (0.73 + 1.82) eV or E = 2.55 eV (b) In case of a hydrogen atom, Since, E1 = - 13.6 eV , E2 = - 3.4 eV , E3 = - 1.5 eV E4 = - 0.85 eV E4 - E2 = 2.55 eV Therefore, quantum numbers of the two levels involved in the emission of these photons are 4 and 2(4 ® 2). (c) Change in angular momentum in transition from 4 to 2 will be DL = L2 - L4 =2 èæç h øö÷ -4 æèç h ÷øö or DL = - h 2p 2p p (d) From conservation of linear momentum |Momentum of hydrogen atom|=|Momentum of emitted photon| or mv = E (m = mass of hydrogen atom) c or v = E = (2.55 ´ 1.6 ´ 10-19 J) m/s) mc (1.67 ´ 10-27 kg) (3.0 ´ 108 v = 0.814 m/s Ans.

300 — Optics and Modern Physics V Example 20 If an X-ray tube operates at the voltage of 10 kV , find the ratio of the de-Broglie wavelength of the incident electrons to the shortest wavelength of X-rays produced. The specific charge of electron is 1.8 ´ 1011C/kg. Solution de-Broglie wavelength when a charge q is accelerated by a potential difference of V volts is lb = h …(i) 2qVm For cut off wavelength of X-rays, we have qV = hc or lm lm = hc …(ii) qV qV From Eqs. (i) and (ii), we get lb = 2m lm c For electron q = 1.8 ´ 1011 C/kg (given). Substituting the values the desired ratio is m 1.8 ´ 1011 ´ 10 ´ 103 lb = 2 = 0.1 Ans. lm 3 ´ 108 V Example 21 The wavelength of the first line of Lyman series for hydrogen is identical to that of the second line of Balmer series for some hydrogen like ion x. Calculate energies of the first four levels of x. Solution Wavelength of the first line of Lyman series for hydrogen atom will be given by the equation 1 = R æçè 1 – 1 ÷øö = 3R …(i) l1 12 22 4 The wavelength of second Balmer line for hydrogen like ion x is 1 2çæ 1 1 öø÷ 3RZ 2 l2 è 22 42 16 = RZ – = …(ii) Given that l1 = l2 or 1 =1 i.e. l1 l2 3R = 3RZ 2 4 16 \\ Z =2 i.e. x ion is He+ . The energies of first four levels of x are E1 = – (13.6) Z 2 = – 54.4 eV E2 = E1 = – 13.6 eV (2)2 E3 = E1 = – 6.04 eV (3)2 and E4 = E1 = – 3.4 eV Ans. (4)2

Chapter 33 Modern Physics - I — 301 V Example 22 A moving hydrogen atom makes a head on collision with a stationary hydrogen atom. Before collision both atoms are in ground state and after collision they move together. What is the minimum value of the kinetic energy of the moving hydrogen atom, such that one of the atoms reaches one of the excited state? Solution Let K be the kinetic energy of the moving hydrogen atom and K ¢, the kinetic energy of combined mass after collision. K m K¢ n=2 m 2m DE = 10.2 eV n=1 From conservation of linear momentum, p = p¢ or 2Km = 2K ¢ (2m) or K = 2K ¢ …(i) …(ii) From conservation of energy, K = K ¢ + DE Solving Eqs. (i) and (ii), we get DE = K Ans. 2 Now, minimum value of DE for hydrogen atom is 10.2 eV. or DE ³ 10.2 eV \\ K ³ 10.2 2 \\ K ³ 20.4 eV Therefore, the minimum kinetic energy of moving hydrogen is 20.4 eV. V Example 23 An imaginary particle has a charge equal to that of an electron and mass 100 times the mass of the electron. It moves in a circular orbit around a nucleus of charge + 4e. Take the mass of the nucleus to be infinite. Assuming that the Bohr model is applicable to this system. (a) Derive an expression for the radius of nth Bohr orbit. (b) Find the wavelength of the radiation emitted when the particle jumps from fourth orbit to the second orbit. m pv2 = 1 Ze2 …(i) Solution (a) We have rn 4pe0 rn2 The quantization of angular momentum gives m pvrn = nh …(ii) 2p Solving Eqs. (i) and (ii), we get r = n2h2e0 Zpm pe2 Substituting mp = 100 m

302 — Optics and Modern Physics where, m = mass of electron and Z = 4 We get, rn = n2h2e0 Ans. 400 p me2 Ans. (b) As we know, E1H = – 13.60 eV and En µ èççæ Z 2 øö÷÷ m n 2 For the given particle, E4 = (–13.60) (4)2 ´ 100 (4)2 = – 1360 eV and E2 = (–13.60) (4)2 ´ 100 (2)2 = – 5440 eV DE = E4 – E2 = 4080 eV \\ l (in Å) = 12375 DE (in eV ) = 12375 4080 = 3.0 Å V Example 24 The energy levels of a hypothetical one electron atom are given by En = - 18.0 eV n2 where n = 1, 2, 3,¼ (a) Compute the four lowest energy levels and construct the energy level diagram. (b) What is the first excitation potential (c) What wavelengths (Å) can be emitted when these atoms in the ground state are bombarded by electrons that have been accelerated through a potential difference of 16.2 V? (d) If these atoms are in the ground state, can they absorb radiation having a wavelength of 2000 Å? (e) What is the photoelectric threshold wavelength of this atom? Solution (a) E1 = –18.0 = – 18.0 eV (1)2 E2 = –18.0 = – 4.5 eV (2)2 E3 = –18.0 = – 2.0 eV (3)2 and E4 = –18.0 = – 1.125 eV (4)2

Chapter 33 Modern Physics - I — 303 The energy level diagram is shown in figure. E4 = –1.125 eV E3 = –2.0 eV E2 = – 4.5 eV E1 = –18.0 eV (b) E2 - E1 = 13.5 eV \\ First excitation potential is 13.5 V. (c) Energy of the electron accelerated by a potential difference of 16.2 V is 16.2 eV. With this energy the electron can excite the atom from n = 1 to n = 3 as E4 – E1 = – 1.125 – (–18.0) = 16.875 eV > 16.2 eV and E3 – E1 = – 2.0 – (–18.0) = 16.0 eV < 16.2 eV Now, l32 = 12375 = – 12375 E3 – E2 2.0 – (– 4.5) = 4950 Å Ans. l31 = 12375 = 12375 = 773 Å Ans. E3 – E1 16 and l21 = 12375 = 12375 E2 – E1 – 4.5 – (–18.0) = 917 Å Ans. (d) No, the energy corresponding to l = 2000 Å is E = 12375 = 6.1875 eV Ans. 2000 The minimum excitation energy is 13.5 eV (n = 1 to n = 2). (e) Threshold wavelength for photoemission to take place from such an atom is lmin = 12375 18 = 687.5 Å Ans. V Example 25 In a photocell the plates P and Q have a separation of 5 cm, which are connected through a galvanometer without any cell. Bichromatic light of wavelengths 4000 Å and 6000 Å are incident on plate Q whose work-function is 2.39 eV . If a uniform magnetic field B exists parallel to the plates, find the minimum value of B for which the galvanometer shows zero deflection. Solution Energy of photons corresponding to light of wavelength l1 = 4000 Å is E1 = 12375 = 3.1 eV 4000 and that corresponding to l2 = 6000 Å is E2 = 12375 = 2.06 eV 6000

304 — Optics and Modern Physics As, E2 < W and E1 > W (W = work-function) Photoelectric emission is possible with l1 only. Gu r P B d Q Photoelectrons experience magnetic force and move along a circular path. The galvanometer will indicate zero deflection if the photoelectrons just complete semicircular path before reaching the plate P. Thus, d = r = 5 cm \\ r = 5 cm = 0.05 m Further, r = mv = 2Km Bq Bq \\ Bmin = 2Km rq Here, K = E1 – W = (3.1 – 2.39) = 0.71 eV Substituting the values, we have Bmin = 2 ´ 0.71 ´ 1.6 ´ 10–19 ´ 9.109 ´ 10–31 (0.05) (1.6 ´ 10–19 ) = 5.68 ´ 10–5 T Ans.

Exercises LEVEL 1 Assertion and Reason Directions : Choose the correct option. (a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion. (b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion. (c) If Assertion is true, but the Reason is false. (d) If Assertion is false but the Reason is true. 1. Assertion : X-rays cannot be deflected by electric or magnetic fields. Reason : These are electromagnetic waves. 2. Assertion : If wavelength of light is doubled, energy and momentum of photons are reduced to half. Reason : By increasing the wavelength, speed of photons will decrease. 3. Assertion : We can increase the saturation current in photoelectric experiment without increasing the intensity of light. Reason : Intensity can be increased by increasing the frequency of incident photons. 4. Assertion : Photoelectric effect proves the particle nature of light. Reason : Photoemission starts as soon as light is incident on the metal surface, provided frequency of incident light is greater than or equal to the threshold frequency. 5. Assertion : During de-excitation from n = 6 to n = 3, total six emission lines may be obtained. Reason : From n = n to n = 1, total n (n - 1) emission lines are obtained. 2 6. Assertion : If frequency of incident light is doubled, the stopping potential will also become two times. Reason : Stopping potential is given by V0 = h (n - n0) e 7. Assertion : X-rays cannot be obtained in the emission spectrum of hydrogen atom. Reason : Maximum energy of photons emitted from hydrogen spectrum is 13.6 eV. 8. Assertion : If applied potential difference in coolidge tube is increased, then difference between K a wavelength and cut off wavelength will increase. Reason : Cut off wavelength is inversely proportional to the applied potential difference in coolidge tube. 9. Assertion : In n = 2, energy of electron in hydrogen like atoms is more compared to n = 1. Reason : Electrostatic potential energy in n = 2 is more. 10. Assertion : In continuous X-ray spectrum, all wavelengths can be obtained. Reason : Accelerated (or retarded) charged particles radiate energy. This is the cause of production of continuous X-rays.

306 — Optics and Modern Physics Objective Questions 1. According to Einstein’s photoelectric equation, the plot of the maximum kinetic energy of the emitted photoelectrons from a metal versus frequency of the incident radiation gives a straight line whose slope (a) depends on the nature of metal used (b) depends on the intensity of radiation (c) depends on both intensity of radiation and the nature of metal used (d) is the same for all metals and independent of the intensity of radiation 2. The velocity of the electron in the first Bohr orbit as compared to that of light is about (a) 1/300 (b) 1/500 (c) 1/137 (d) 1/187 3. 86A222 ® 84B210. In this reaction, how many a and b particles are emitted? (a) 6 a ,3 b (b) 3 a ,4 b (c) 4 a ,3 b (d) 3 a ,6 b 4. An X-ray tube is operated at 20 kV. The cut off wavelength is (a) 0.89 Å (b) 0.75 Å (c) 0.62 Å (d) None of these 5. An X-ray tube is operated at 18 kV. The maximum velocity of electron striking the target is (a) 8 ´107 m/s (b) 6 ´107 m/s (c) 5 ´107 m/s (d) None of these 6. What is the ratio of de-Broglie wavelength of electron in the second and third Bohr orbits in the hydrogen atoms? (a) 2/3 (b) 3/2 (c) 4/3 (d) 3/4 7. The energy of a hydrogen like atom (or ion) in its ground state is – 122.4 eV. It may be (a) hydrogen atom (b) He+ (c) Li2+ (d) Be3 + 8. The operating potential in an X-ray tube is increased by 2%. The percentage change in the cut off wavelength is (a) 1% increase (b) 2% increase (c) 2% decrease (d) 1% decrease 9. The energy of an atom or ion in the first excited state is –13.6 eV. It may be (a) He+ (b) Li+ + (c) hydrogen (d) deuterium 10. In order that the short wavelength limit of the continuous X-ray spectrum be 1 Å, the potential difference through which an electron must be accelerated is (a) 124 kV (b) 1.24 kV (c) 12.4 kV (d) 1240 kV 11. The momentum of an X-ray photon with l = 0.5 Å is (a) 13.26 ´10-26 kg-m/s (b) 1.326 ´10-26 kg-m/s (c) 13.26 ´10-24 kg-m/s (d) 13.26 ´10-22 kg-m/s

Chapter 33 Modern Physics - I — 307 12. The work-function of a substance is 1.6 eV. The longest wavelength of light that can produce photoemission from the substance is (a) 7750 Å (b) 3875 Å (c) 5800 Å (d) 2900 Å 13. Find the binding energy of an electron in the ground state of a hydrogen like atom in whose spectrum the third Balmer line is equal to 108.5 nm. (a) 54.4 eV (b) 13.6 eV (c) 112.4 eV (d) None of these 14. Let the potential energy of hydrogen atom in the ground state be zero. Then, its energy in the first excited state will be (a) 10.2 eV (b) 13.6 eV (c) 23.8 eV (d) 27.2 eV 15. Light of wavelength 330 nm falling on a piece of metal ejects electrons with sufficient energy with required voltage V0 to prevent them from reaching a collector. In the same set up, light of wavelength 220 nm ejects electrons which require twice the voltage V0 to stop them in reaching a collector. The numerical value of voltage V0 is (a) 16 V (b) 15 V 15 16 (c) 15 V (d) 8 V 8 15 16. Maximum kinetic energy of a photoelectron is E when the wavelength of incident light is l. If energy becomes four times when wavelength is reduced to one-third, then work-function of the metal is (a) 3 hc (b) hc l 3l (c) hc (d) hc l 2l 17. If the frequency of K a X-ray emitted from the element with atomic number 31 is f, then the frequency of K a X-ray emitted from the element with atomic number 51 would be (a) 5 f (b) 51 f 3 31 (c) 9 f (d) 25 f 25 9 18. According to Moseley’s law, the ratio of the slope of graph between f and Z for K b and K a is (a) 32 (b) 27 27 32 (c) 5 (d) 36 36 5 19. If the electron in hydrogen orbit jumps from third orbit to second orbit, the wavelength of the emitted radiation is given by (a) l = R (b) l = 5 6 R (c) l = 36 (d) l = 5R 5R 36

308 — Optics and Modern Physics 20. A potential of 10000 V is applied across an X-ray tube. Find the ratio of de-Broglie wavelength associated with incident electrons to the minimum wavelength associated with X-rays. (Given, e/m =1.8 ´ 1011 C/kg for electrons) (a) 10 (b) 20 (c) 1/10 (d) 1/20 21. When a metallic surface is illuminated with monochromatic light of wavelength l, the stopping potential is 5 V0. When the same surface is illuminated with the light of wavelength 3l, the stopping potential is V0. Then, the work-function of the metallic surface is (a) hc/6l (b) hc/5l (c) hc/4l (d) 2hc/4l 22. The threshold frequency for a certain photosensitive metal is n0. When it is illuminated by light of frequency n = 2n0, the stopping potential for photoelectric current is V0. What will be the stopping potential when the same metal is illuminated by light of frequency n = 3n0? (a) 1.5 V0 (b) 2 V0 (c) 2.5 V0 (d) 3 V0 23. The frequency of the first line in Lyman series in the hydrogen spectrum is n. What is the frequency of the corresponding line in the spectrum of doubly ionized Lithium? (a) n (b) 3 n (c) 9 n (d) 2 n 24. Which energy state of doubly ionized lithium (Li++) has the same energy as that of the ground state of hydrogen? (a) n = 1 (b) n = 2 (c) n = 3 (d) n = 4 25. Two identical photo-cathodes receive light of frequencies n1 and n2. If the velocities of the photoelectrons (of mass m) coming out are v1 and v2 respectively, then - n2)ùûú1/2 (a) v1 - v2 = é 2h (n1 (b) v12 - v22 = 2h (n1 - n2) êë m m (c) v1 + v2 = é 2h (n1 - n2)úûù1/2 (d) v12 + v22 = 2h (n1 - n2) êë m m 26. The longest wavelength of the Lyman series for hydrogen atom is the same as the wavelength of a certain line in the spectrum of He+ when the electron makes a transition from n ® 2. The value of n is (a) 3 (b) 4 (c) 5 (d) 6 27. The wavelength of the K a - line for the uranium is (Z = 92) (R = 1.0973 ´107 m-1) (a) 1.5 Å (b) 0.5 Å (c) 0.15 Å (d) 2.0 Å 28. The frequencies of K a , K b and La X-rays of a material are g1, g2 and g3 respectively. Which of the following relation holds good? (a) g2 = g1g+1 gg33 (b) g2 = g1 + g3 (c) g2 = 2 (d) g3 = g1 g2 29. A proton and an a-particle are accelerated through same potential difference. Then, the ratio of de-Broglie wavelength of proton and a-particle is (a) 2 (b) 1 (c) 2 2 (d) None of these 2

Chapter 33 Modern Physics - I — 309 30. If E1, E2 and E3 represent respectively the kinetic energies of an electron , an a-particle and a proton each having same de-Broglie wavelength, then (a) E1 > E3 > E2 (b) E2 > E3 > E1 (c) E1 > E2 > E3 (d) E1 = E2 = E3 31. If the potential energy of a hydrogen atom in the ground state is assumed to be zero, then total energy of n = ¥ is equal to (a) 13.6 eV (b) 27.2 eV (c) zero (d) None of these 32. A 1000 W transmitter works at a frequency of 880 kHz. The number of photons emitted per second is (a) 1.7 ´1028 (b) 1.7 ´1030 (c) 1.7 ´1023 (d) 1.7 ´1025 33. Electromagnetic radiation of wavelength 3000 Å is incident on an isolated platinum surface of work-function 6.30 eV. Due to the radiation, the (a) sphere becomes positively charged (b) sphere becomes negatively charged (c) sphere remains neutral (d) maximum kinetic energy of the ejected photoelectrons would be 2.03 eV 34. The energy of a hydrogen atom in its ground state is – 13.6 eV. The energy of the level corresponding to the quantum number n = 5 is (a) – 0.54 eV (b) – 5.40 eV (c) – 0.85 eV (d) – 2.72 eV 35. Ultraviolet radiation of 6.2 eV falls on an aluminium surface (work-function = 4.2 eV). The kinetic energy in joule of the fastest electrons emitted is (a) 3.2 ´ 10-21 (b) 3.2 ´ 10-19 (c) 3.2 ´ 10-17 (d) 3.2 ´ 10-15 36. What should be the velocity of an electron so that its momentum becomes equal to that of a photon of wavelength 5200 Å? (a) 700 m/s (b) 1000 m/s (c) 1400 m/s (d) 2800 m/s 37. Photoelectric work-function of a metal is 1 eV. Light of wavelength l = 3000 Å falls on it. The photoelectrons come out with maximum velocity (a) 10 m/s (b) 103 m/s (c) 104 m/s (d) 106 m/s Subjective Questions Note You can take approximations in the answers. h = 6.62 ´ 10-34 J-s, c = 3.0 ´ 108 m/ s, me = 9.1 ´ 10-31 kg and 1 eV = 1.6 ´ 10-19 J 1. For a given element the wavelength of the K a -line is 0.71 nm and of the K b-line it is 0.63 nm. Use this information to find wavelength of the La -line. 2. The energy of the n = 2 state in a given element is E2 = – 2870 eV. Given that the wavelengths of the K a and K b-lines are 0.71 nm and 0.63 nm respectively, determine the energies E1 and E3. 3. 1.5 mW of 400 nm light is directed at a photoelectric cell. If 0.1% of the incident photons produce photoelectrons, find the current in the cell.

310 — Optics and Modern Physics 4. A photon has momentum of magnitude 8.24 ´ 10-28 kg-m/ s. (a) What is the energy of this photon? Give your answer in joules and in electron volts. (b) What is the wavelength of this photon? In what region of the electromagnetic spectrum does it lie? 5. A 75 W light source emits light of wavelength 600 nm. (a) Calculate the frequency of the emitted light. (b) How many photons per second does the source emit? 6. An excited nucleus emits a gamma-ray photon with energy of 2.45 MeV. (a) What is the photon frequency? (b) What is the photon wavelength? 7. (a) A proton is moving at a speed much less than the speed of light. It has kinetic energy K1 and momentum p1. If the momentum of the proton is doubled, so p2 = 2 p1, how is its new kinetic energy K 2 related to K1? (b) A photon with energy E1 has momentum p1. If another photon has momentum p2 that is twice p1 , how is the energy E2 of the second photon related to E1? 8. A parallel beam of monochromatic light of wavelength 500 nm is incident normally on a perfectly absorbing surface. The power through any cross-section of the beam is 10 W. Find (a) the number of photons absorbed per second by the surface and (b) the force exerted by the light beam on the surface. 9. A beam of white light is incident normally on a plane surface absorbing 70% of the light and reflecting the rest. If the incident beam carries 10 W of power, find the force exerted by it on the surface. 10. A parallel beam of monochromatic light of wavelength 663 nm is incident on a totally reflecting plane mirror. The angle of incidence is 60° and the number of photons striking the mirror per second is 1.0 ´ 1019. Calculate the force exerted by the light beam on the mirror. 11. Wavelength of Bullet. Calculate the de-Broglie wavelength of a 5.00 g bullet that is moving at 340 m/ s. Will it exhibit wave like properties? 12. (a) An electron moves with a speed of 4.70 ´ 106 m/ s. What is its de-Broglie wavelength? (b) A proton moves with the same speed. Determine its de-Broglie wavelength. 13. An electron has a de-Broglie wavelength of 2.80 ´ 10-10 m. Determine (a) the magnitude of its momentum, (b) its kinetic energy (in joule and in electron volt). 14. Find de-Broglie wavelength corresponding to the root-mean square velocity of hydrogen molecules at room temperature (20°C). 15. An electron, in a hydrogen like atom, is in excited state. It has a total energy of –3.4 eV, find the de-Broglie wavelength of the electron. 16. In the Bohr model of the hydrogen atom, what is the de-Broglie wavelength for the electron when it is in (a) the n = 1 level? (b) the n = 4 level? In each case, compare the de-Broglie wavelength to the circumference 2prn of the orbit. 17. The binding energy of an electron in the ground state of He atom is equal to E0 = 24.6 eV. Find the energy required to remove both electrons from the atom.

Chapter 33 Modern Physics - I — 311 18. Hydrogen atom in its ground state is excited by means of monochromatic radiation of wavelength 1023 Å. How many different lines are possible in the resulting spectrum? Calculate the longest wavelength among them. You may assume the ionization energy of hydrogen atom as 13.6 eV. 19. A doubly ionized lithium atom is hydrogen like with atomic number 3. Find the wavelength of the radiation required to excite the electron in Li++ from the first to the third Bohr orbit (ionization energy of the hydrogen atom equals 13.6 eV). 20. Find the quantum number n corresponding to nth excited state of He+ ion if on transition to the ground state the ion emits two photons in succession with wavelengths 108.5 nm and 30.4 nm. The ionization energy of the hydrogen atom is 13.6 eV. 21. A hydrogen like atom (described by the Bohr model) is observed to emit ten wavelengths, originating from all possible transitions between a group of levels. These levels have energies between –0.85 eV and –0.544 eV (including both these values). (a) Find the atomic number of the atom. (b) Calculate the smallest wavelength emitted in these transitions. (Take ground state energy of hydrogen atom = - 13.6 eV) 22. The energy levels of a hypothetical one electron atom are µ 0 eV –0.80 eV shown in the figure. n=5 –1.45 eV –3.08 eV (a) Find the ionization potential of this atom. n=4 –5.30 eV (b) Find the short wavelength limit of the series terminating at n = 3 –15.6 eV n = 2. n=2 (c) Find the excitation potential for the state n = 3. n=1 (d) Find wave number of the photon emitted for the transition n = 3 to n = 1. 23. (a) An atom initially in an energy level with E = – 6.52 eV absorbs a photon that has wavelength 860 nm. What is the internal energy of the atom after it absorbs the photon? (b) An atom initially in an energy level with E = - 2.68 eV emits a photon that has wavelength 420 nm. What is the internal energy of the atom after it emits the photon? 24. A silver ball is suspended by a string in a vacuum chamber and ultraviolet light of wavelength 2000 Å is directed at it. What electrical potential will the ball acquire as a result? Work function of silver is 4.3 eV. 25. A small particle of mass m moves in such a way that the potential energy U = 1 m2w2r 2, 2 where wis a constant and r is the distance of the particle from the origin. Assuming Bohr model of quantization of angular momentum and circular orbits, show that radius of the nth allowed orbit is proportional to n. 26. Wavelength of K a -line of an element is l0. Find wavelength of K b-line for the same element. 27. X-rays are produced in an X-ray tube by electrons accelerated through an electric potential difference of 50.0 kV. An electron makes three collisions in the target coming to rest and loses half its remaining kinetic energy in each of the first two collisions. Determine the wavelength of the resulting photons. (Neglecting the recoil of the heavy target atoms). 28. From what material is the anode of an X-ray tube made, if the K a - line wavelength of the characteristic spectrum is 0.76 Å? 29. A voltage applied to an X-ray tube being increased h = 1.5 times, the short wave limit of an X-ray continuous spectrum shifts by Dl = 26 pm. Find the initial voltage applied to the tube.

312 — Optics and Modern Physics 30. The K a X-rays of aluminium (Z = 13) and zinc (Z = 30) have wavelengths 887 pm and 146 pm, respectively. Use Moseley’s equation n = a(Z - b) to find the wavelength of the K a X-ray of iron (Z = 26). 31. Characteristic X-rays of frequency 4.2 ´ 1018 Hz are produced when transitions from L shell take place in a certain target material. Use Moseley’s law and determine the atomic number of the target material. Given, Rydberg constant is R = 1.1 ´ 107 m-1. 32. The electric current in an X-ray tube operating at 40 kV is 10 mA. Assume that on an average 1% of the total kinetic energy of the electrons hitting the target are converted into X-rays. (a) What is the total power emitted as X-rays and (b) How much heat is produced in the target every second? 33. The stopping potential for the photoelectrons emitted from a metal surface of work-function 1.7 eV is 10.4 V. Find the wavelength of the radiation used. Also, identify the energy levels in hydrogen atom, which will emit this wavelength. 34. What will be the maximum kinetic energy of the photoelectrons ejected from magnesium (for which the work-function W = 3.7 eV) when irradiated by ultraviolet light of frequency 1.5 ´ 1015 s-1. 35. A metallic surface is irradiated with monochromatic light of variable wavelength. Above a wavelength of 5000 Å, no photoelectrons are emitted from the surface. With an unknown wavelength, stopping potential of 3 V is necessary to eliminate the photocurrent. Find the unknown wavelength. 36. A graph regarding photoelectric effect is shown between the maximum kinetic energy of electrons and the frequency of the incident light . On the basis of data as shown in the graph, calculate Kmax (eV) 8 6 4 2 D 0A 10 20 30 –2 f (´ 1014 Hz) –4 C (a) threshold frequency, (b) work-function, (c) planck constant 37. A metallic surface is illuminated alternatively with light of wavelengths 3000 Å and 6000 Å. It is observed that the maximum speeds of the photoelectrons under these illuminations are in the ratio 3 : 1. Calculate the work-function of the metal and the maximum speed of the photoelectrons in two cases. 38. Light of wavelength 180 nm ejects photoelectrons from a plate of metal whose work-function is 2 eV. If a uniform magnetic field of 5 ´ 10–5 T be applied parallel to the plate, what would be the radius of the path followed by electrons ejected normally from the plate with maximum energy. 39. Light described at a place by the equation E = (100 V/ m) [sin(5 ´ 1015 s–1)t + sin(8 ´ 1015 s–1)t] falls on a metal surface having work-function 2.0 eV. Calculate the maximum kinetic energy of the photoelectrons. 40. The electric field associated with a light wave is given by E = E0 sin [(1.57 ´ 107 m-1) (x - ct)]. Find the stopping potential when this light is used in an experiment on photoelectric effect with a metal having work-function 1.9 eV.

LEVEL 2 Single Correct Option 1. If we assume only gravitational attraction between proton and electron in hydrogen atom and the Bohr quantization rule to be followed, then the expression for the ground state energy of the atom will be (the mass of proton is M and that of electron is m.) (a) G2M 2m2 (b) - 2p2G2M 2m3 h2 h2 (c) - 2p2GM 2m3 (d) None of these h2 2. An electron in a hydrogen atom makes a transition from first excited state to ground state. The magnetic moment due to circulating electron (a) increases two times (b) decreases two times (c) increases four times (d) remains same 3. The excitation energy of a hydrogen like ion to its first excited state is 40.8 eV. The energy needed to remove the electron from the ion in the ground state is (a) 54.4 eV (b) 62.6 eV (c) 72.6 eV (d) 58.6 eV 4. An electron in a hydrogen atom makes a transition from first excited state to ground state. The equivalent current due to circulating electron (a) increases 4 times (b) decreases 4 times (c) increases 8 times (d) decreases 8 times 5. In a sample of hydrogen like atoms all of which are in ground state, a photon beam containing photons of various energies is passed. In absorption spectrum, five dark lines are observed. The number of bright lines in the emission spectrum will be (assume that all transitions take place) (a) 21 (b) 10 (c) 15 (d) None of these 6. Let An be the area enclosed by the nth orbit in a hydrogen atom. The graph of ln ( An / A1) against ln (n) (a) will not pass through origin (b) will be a straight line with slope 4 (c) will be a rectangular hyperbola (d) will be a parabola 7. In the hydrogen atom, an electron makes a transition from n = 2 to n = 1. The magnetic field produced by the circulating electron at the nucleus (a) decreases 16 times (b) increases 4 times (c) decreases 4 times (d) increases 32 times 8. A stationary hydrogen atom emits photon corresponding to the first line of Lyman series. If R is the Rydberg constant and M is the mass of the atom, then the velocity acquired by the atom is (a) 3Rh (b) 4M 4M 3Rh (c) Rh (d) 4M 4M Rh

314 — Optics and Modern Physics 9. Light wave described by the equation 200 V/m sin (1.5 ´ 1015 s-1) t cos (0.5 ´ 1015 s-1) t falls on metal surface having work-function 2.0 eV. Then, the maximum kinetic energy of photoelectrons is (a) 3.27 eV (b) 2.2 eV (c) 2.85 eV (d) None of these 10. A hydrogen like atom is excited using a radiation. Consequently, six spectral lines are observed in the spectrum. The wavelength of emission radiation is found to be equal or smaller than the radiation used for excitation. This concludes that the gas was initially at (a) ground state (b) first excited state (c) second excited state (d) third excited state 11. The time period of the electron in the ground state of hydrogen atom is two times the time period of the electron in the first excited state of a certain hydrogen like atom (Atomic number Z). The value of Z is (a) 2 (b) 3 (c) 4 (d) None of these 12. The wavelengths of K a X-rays from lead isotopes Pb204, Pb206 and Pb208 are l1, l2 and l3 respectively. Choose the correct alternative. (a) l1 < l2 < l3 (b) l1 > l2 > l3 (c) l1 = l2 = l3 (d) None of these 13. In case of hydrogen atom, whenever a photon is emitted in the Balmer series, (a) there is a probability of emitting another photon in the Lyman series (b) there is a probability of emitting another photon of wavelength 1213 Å (c) the wavelength of radiation emitted in Lyman series is always shorter than the wavelength emitted in the Balmer series (d) All of the above 14. An electron of kinetic energy K collides elastically with a stationary hydrogen atom in the ground state. Then, (a) K > 13.6 eV (b) K > 10.2 eV (c) K < 10.2 eV (d) data insufficient 15. In a stationary hydrogen atom, an electron jumps from n = 3 to n = 1. The recoil speed of the hydrogen atom is about (a) 4 m/s (b) 4 cm/s (c) 4 mm/s (d) 4 ´ 10-4 m/s 16. An X-ray tube is operating at 150 kV and 10 mA. If only 1% of the electric power supplied is converted into X-rays, the rate at which the target is heated in calorie per second is (a) 3.55 (b) 35.5 (c) 355 (d) 3550 17. An electron revolves round a nucleus of atomic number Z. If 32.4 eV of energy is required to excite an electron from the n = 3 state to n = 4 state, then the value of Z is (a) 5 (b) 6 (c) 4 (d) 7 18. If the de-Broglie wavelength of a proton is 10-13 m, the electric potential through which it must have been accelerated is (a) 4.07 ´ 104 V (b) 8.15 ´ 104 V (c) 8.15 ´ 103 V (d) 4.07 ´ 105 V

Chapter 33 Modern Physics - I — 315 19. If En and Ln denote the total energy and the angular momentum of an electron in the nth orbit of Bohr atom, then (a) En µ Ln (b) En µ 1 (c) En µ L2n Ln (d) En µ1 L2n 20. An orbital electron in the ground state of hydrogen has the magnetic moment m1. This orbital electron is excited to 3rd excited state by some energy transfer to the hydrogen atom. The new magnetic moment of the electron is m 2, then (a) m1 = 4m 2 (b) 2 m1 = m 2 (c) 16 m1 = m 2 (d) 4 m1 = m 2 21. A moving hydrogen atom makes a head-on collision with a stationary hydrogen atom. Before collision, both atoms are in ground state and after collision they move together. The minimum value of the kinetic energy of the moving hydrogen atom, such that one of the atoms reaches one of the excitation state is (a) 20.4 eV (b) 10.2 eV (c) 54.4 eV (d) 13.6 eV 22. In an excited state of hydrogen like atom an electron has total energy of – 3.4 eV. If the kinetic energy of the electron is E and its de-Broglie wavelength is l, then (a) l = 6.6 Å (b) E = 3.4 eV (c) Both are correct (d) Both are wrong More than One Correct Options 1. If the potential difference of coolidge tube producing X-ray is increased, then choose the correct option (s). (a) the interval between lKa and lKb increases (b) the interval between lKa and l0 increases (c) the interval between lKb and l0 increases (d) l0 does not change Here, l0 is cut off wavelength and lK a and lK b are wavelengths of K a and K b characteristic X-rays. 2. In Bohr model of the hydrogen atom, let R, v and E represent the radius of the orbit, speed of the electron and the total energy of the electron respectively. Which of the following quantities are directly proportional to the quantum number n? (a) vR (b) RE (c) v (d) R E E 3. The magnitude of angular momentum, orbital radius and time period of revolution of an electron in a hydrogen atom corresponding to the quantum number n are L, r and T respectively. Which of the following statement(s) is/are correct? (a) rL is independent of n (b) L µ1 T T n2 (c) T µ n (d) Lr µ 1 r n3

316 — Optics and Modern Physics 4. In which of the following cases the heavier of the two particles has a smaller de-Broglie wavelength? The two particles (a) move with the same speed (b) move with the same linear momentum (c) move with the same kinetic energy (d) have the same change of potential energy in a conservative field 5. Hydrogen atom absorbs radiations of wavelength l0 and consequently emit radiations of 6 different wavelengths, of which two wavelengths are longer than l0. Choose the correct alternative(s). (a) The final excited state of the atoms is n = 4 (b) The initial state of the atoms is n = 2 (c) The initial state of the atoms is n = 3 (d) There are three transitions belonging to Lyman series 6. In coolidge tube, if f and l represent the frequency and wavelength of K a -line for a metal of atomic number Z, then identify the statement which represents a straight line (a) f versus Z (b) 1 versus Z l (c) f versus Z (d) l versus Z Comprehension Based Questions Passage I (Q. No. 1 to 3) When a surface is irradiated with light of wavelength 4950 Å, a photocurrent appears which vanishes if a retarding potential greater than 0.6 volt is applied across the phototube. When a second source of light is used, it is found that the critical retarding potential is changed to 1.1 volt. 1. The work-function of the emitting surface is (a) 2.2 eV (b) 1.5 eV (c) 1.9 eV (d) 1.1 eV 2. The wavelength of the second source is (a) 6150 Å (b) 5150 Å (c) 4125 Å (d) 4500 Å 3. If the photoelectrons (after emission from the source) are subjected to a magnetic field of 10 tesla, the two retarding potentials would (a) uniformly increase (b) uniformly decrease (c) remain the same (d) None of these Passage II (Q. No. 4 to 6) In an experimental set up to study the photoelectric effect a point source of light of power 3.2 ´ 10-3 W was taken. The source can emit monoenergetic photons of energy 5 eV and is located at a distance of 0.8 m from the centre of a stationary metallic sphere of work-function 3.0 eV . The radius of the sphere is r = 8 ´ 10-3 m. The efficiency of photoelectric emission is one for every 106 incident photons. Based on the information given above answer the questions given below. (Assume that the sphere is isolated and photoelectrons are instantly swept away after the emission).

Chapter 33 Modern Physics - I — 317 4. de-Broglie wavelength of the fastest moving photoelectron is (a) 6.63 Å (b) 8.69 Å (c) 2 Å (d) 5.26 Å 5. It was observed that after some time emission of photoelectrons from the sphere stopped. Charge on the sphere when the photon emission stops is (a) 16pe0r coulomb (b) 8pe0r coulomb (c) 15pe0r coulomb (d) 20pe0r coulomb 6. Time after which photoelectric emission stops is (a) 100 s (b) 121 s (c) 111 s (d) 141 s Match the Columns 1. Match the following two columns for hydrogen spectrum. Column I Column II (a) Lyman series (p) infrared region (b) Balmer series (q) visible region (c) Paschen series (r) ultraviolet region (d) Brackett series (s) X-rays 2. Ionization energy from first excited state of hydrogen atom is E. Match the following two columns for He+ atom. Column I Column II (a) Ionization energy from (p) 4 E ground state (b) Electrostatic potential (q) – 16 E energy in first excited state. (c) Kinetic energy of electron in (r) – 8 E ground state. (d) Ionization energy from first (s) 16 E excited state. 3. Kmax v0 2 1 ff –Y1 –Y2 Maximum kinetic energy versus frequency of incident light and stopping potential versus frequency of incident light graphs are shown in figure. Match the following two columns.

318 — Optics and Modern Physics Column I Column II (a) Slope of line-1 (p) h/e (b) Slope of line-2 (q) h (c) Y1 (r) W (d) Y2 (s) W/e Here, h = Planck constant, e = 1.6 ´ 10-19 C and W = work-function. 4. For hydrogen and hydrogen type atoms, match the following two columns. Column I Column II (a) Time period (p) Proportional to n/Z (b) Angular momentum (q) Proportional to n2/Z (c) Speed (r) Proportional to n3 /Z 2 (d) Radius (s) None of these 5. In hydrogen atom wavelength of second line of Balmer series is l. Match the following two columns corresponding to the wavelength. Column I Column II (a) First line of Balmer series (p) (27/20) l (b) Third line of Balmer series (q) (l/4) (c) First line of Lyman series (r) (25/12) l (d) Second line of Lyman series (s) None of these 6. Match the following (Give most appropriate one matching) Column I Column II (a) Characteristic X-ray (p) Inverse process of photoelectric effect (b) X-ray production (c) Cut off wavelength (q) Potential difference (d) Continuous X-ray (r) Moseley’s law (s) None of these 7. In a photoelectric effect experiment. If f is the frequency of radiations incident on the metal surface and I is the intensity of the incident radiations, then match the following columns. Column I Column II (p) Stopping potential increases (a) If f is increased keeping I and work-function constant (q) Saturation current increases (b) If distance between cathode (r) Maximum kinetic energy of and anode is increased photoelectron increases (c) If I is increased keeping f and (s) Stopping potential remains same work-function constant (d) Work-function is decreased keeping f and I constant

Chapter 33 Modern Physics - I — 319 Subjective Questions 1. The wavelength for n = 3 to n = 2 transition of the hydrogen atom is 656.3 nm. What are the wavelengths for this same transition in (a) positronium, which consists of an electron and a positron (b) singly ionized helium (Note: A positron is a positively charged electron). 2. (a) Find the frequencies of revolution of electrons in n = 1 and n = 2 Bohr orbits. (b) What is the frequency of the photon emitted when an electron in an n = 2 orbit drops to an n = 1 hydrogen orbit? (c) An electron typically spends about 10–8s in an excited state before it drops to a lower state by emitting a photon. How many revolutions does an electron in an n = 2 Bohr hydrogen orbit make in 1.00 ´ 10-8 s? 3. A muon is an unstable elementary particle whose mass is 207 me and whose charge is either + e or – e. A negative muon (m – ) can be captured by a nucleus to form a muonic atom. (a) A proton captures a m – . Find the radius of the first Bohr orbit of this atom. (b) Find the ionization energy of the atom. 4. (a) A gas of hydrogen atoms in their ground state is bombarded by electrons with kinetic energy 12.5 eV. What emitted wavelengths would you expect to see? (b) What if the electrons were replaced by photons of same energy? 5. A source emits monochromatic light of frequency 5.5 ´ 1014 Hz at a rate of 0.1 W. Of the photons given out, 0.15% fall on the cathode of a photocell which gives a current of 6 mA in an external circuit. (a) Find the energy of a photon. (b) Find the number of photons leaving the source per second. (c) Find the percentage of the photons falling on the cathode which produce photoelectrons. 6. The hydrogen atom in its ground state is excited by means of monochromatic radiation. Its resulting spectrum has six different lines. These radiations are incident on a metal plate. It is observed that only two of them are responsible for photoelectric effect. If the ratio of maximum kinetic energy of photoelectrons in the two cases is 5 then find the work-function of the metal. 7. Electrons in hydrogen like atoms (Z = 3) make transitions from the fifth to the fourth orbit and from the fourth to the third orbit. The resulting radiations are incident normally on a metal plate and eject photoelectrons. The stopping potential for the photoelectrons ejected by the shorter wavelength is 3.95 volts. Calculate the work-function of the metal and the stopping potential for the photoelectrons ejected by the longer wavelength. 8. Find an expression for the magnetic dipole moment and magnetic field induction at the centre of Bohr’s hypothetical hydrogen atom in the n th orbit of the electron in terms of universal constant. 9. An electron and a proton are separated by a large distance and the electron approaches the proton with a kinetic energy of 2 eV. If the electron is captured by the proton to form a hydrogen atom in the ground state, what wavelength photon would be given off? 10. Hydrogen gas in the atomic state is excited to an energy level such that the electrostatic potential energy of H-atom becomes –1.7 eV. Now, a photoelectric plate having work-function W = 2.3 eV is exposed to the emission spectra of this gas. Assuming all the transitions to be possible, find the minimum de-Broglie wavelength of the ejected photoelectrons.

320 — Optics and Modern Physics 11. A gas of hydrogen like atoms can absorb radiation of 68 eV. Consequently, the atom emits radiations of only three different wavelengths. All the wavelengths are equal or smaller than that of the absorbed photon. (a) Determine the initial state of the gas atoms. (b) Identify the gas atoms. (c) Find the minimum wavelength of the emitted radiations. (d) Find the ionization energy and the respective wavelength for the gas atoms. 12. A photon with energy of 4.9 eV ejects photoelectrons from tungsten. When the ejected electron enters a constant magnetic field of strength B = 2.5 mT at an angle of 60° with the field direction, the maximum pitch of the helix described by the electron is found to be 2.7 mm. Find the work-function of the metal in electron-volt. Given that specific charge of electron is 1.76 ´ 1011 C/ kg. 13. For a certain hypothetical one-electron atom, the wavelength (in Å) for the spectral lines for transitions originating at n = p and terminating at n = 1 are given by l = 1500 p2 , where p = 2, 3, 4 p2 - 1 (a) Find the wavelength of the least energetic and the most energetic photons in this series. (b) Construct an energy level diagram for this element showing the energies of the lowest three levels. (c) What is the ionization potential of this element? 14. A photocell is operating in saturation mode with a photocurrent 4.8 mA when a monochromatic radiation of wavelength 3000 Å and power of 1 mW is incident. When another monochromatic radiation of wavelength 1650 Å and power 5 mW is incident, it is observed that maximum velocity of photoelectron increases to two times. Assuming efficiency of photoelectron generation per incident photon to be same for both the cases, calculate (a) the threshold wavelength for the cell (b) the saturation current in second case (c) the efficiency of photoelectron generation per incident photon 15. Wavelengths belonging to Balmer series for hydrogen atom lying in the range of 450 nm to 750 nm were used to eject photoelectrons from a metal surface whose work-function is 2.0 eV. Find (in eV) the maximum kinetic energy of the emitted photoelectrons. 16. Assume that the de-Broglie wave associated with an electron can form a standing wave between the atoms arranged in a one-dimensional array with nodes at each of the atomic sites. It is found that one such standing wave is formed if the distance d between the atoms of the array is 2 Å. A similar standing wave is again formed if d is increased to 2.5 Å but not for any intermediate value of d. Find the energy of the electron in eV and the least value of d for which the standing wave of the type described above can form. 17. The negative muon has a charge equal to that of an electron but a mass that is 207 times as great. Consider hydrogen like atom consisting of a proton and a muon. (a) What is the reduced mass of the atom? (b) What is the ground-level energy (in eV)? (c) What is the wavelength of the radiation emitted in the transition from the n = 2 level to the n = 1 level?

Chapter 33 Modern Physics - I — 321 18. Assume a hypothetical hydrogen atom in which the potential energy between electron and proton at separation r is given by U = [k ln r – (k/ 2)], where k is a constant. For such a hypothetical hydrogen atom, calculate the radius of nth Bohr orbit and energy levels. 19. An electron is orbiting in a circular orbit of radius r under the influence of a constant magnetic field of strength B. Assuming that Bohr postulate regarding the quantisation of angular momentum holds good for this electron, find (a) the allowed values of the radius r of the orbit. (b) the kinetic energy of the electron in orbit (c) the potential energy of interaction between the magnetic moment of the orbital current due to the electron moving in its orbit and the magnetic field B. (d) the total energy of the allowed energy levels. (e) the total magnetic flux due to the magnetic field B passing through the nth orbit. (Assume that the charge on the electron is – e and the mass of the electron is m). 20. A mixture of hydrogen atoms (in their ground state) and hydrogen like ions (in their first excited state) are being excited by electrons which have been accelerated by same potential difference V volts. After excitation when they come directly into ground state, the wavelengths of emitted light are found in the ratio 5 : 1. Then, find (a) the minimum value of V for which both the atoms get excited after collision with electrons. (b) atomic number of other ion. (c) the energy of emitted light. 21. When a surface is irradiated with light of l = 4950 Å a photocurrent appears which vanishes if a retarding potential 0.6 V is applied. When a different source of light is used, it is found that critical retarding potential is changed to 1.1 volt. Find the work-function of emitting surface and wavelength of second source. If photoelectrons after emission from surface are subjected to a magnetic field of 10 tesla, what changes will be observed in the above two retarding potentials? 22. In an experiment on photoelectric effect light of wavelength 400 nm is incident on a metal plate at the rate of 5 W. The potential of the collector plate is made sufficiently positive with respect to emitter so that the current reaches the saturation value. Assuming that on the average one out of every 106 photons is able to eject a photoelectron, find the photocurrent in the circuit. 23. A light beam of wavelength 400 nm is incident on a metal of work-function 2.2 eV. A particular electron absorbs a photon and makes 2 collisions before coming out of the metal (a) Assuming that 10% of existing energy is lost to the metal in each collision find the final kinetic energy of this electron as it comes out of the metal. (b) Under the same assumptions find the maximum number of collisions, the electron should suffer before it becomes unable to come out of the metal.

Answers Introductory Exercise 33.1 2. 4.82 ´ 1016 per m2-s 3. 2 1. 4.6 eV, 2.45 ´ 10-27 kg-m 5. (c) 6. (a) 4.81 ´ 10-34 m (b) 7.12 ´ 10-11m s 4. 2 Introductory Exercise 33.2 1. 122.4 eV 2. 3.16 ´ 10–34 kg-m2 /s 3. (d) 4. (a) 5. (b) 11. 3f , f 6. (c) 7. (b) 8. (d) 9. (d) 10. (d) 44 12. 651 nm 6. 42 Introductory Exercise 33.3 4. (a) 1. (b) 2. (b) 3. (a) 4. (a) 5. (b) Introductory Exercise 33.4 2. Kmax µ (f - f0 ) 3. 1.16 ´ 1015 Hz 1. Zero, 3.19 eV 6. (c) 7. (b) 8. (a) 5. (a) Exercises LEVEL 1 Assertion and Reason 5. (a or b) 6. (d) 7. (a) 8. (b) 9. (b) 10. (d) 1. (a) 2. (c) 3. (a or b) 4. (a) Objective Questions 1. (d) 2. (c) 3. (b) 4. (c) 5. (a) 6. (a) 7. (c) 8. (c) 9. (a) 10. (c) 11. (c) 12. (a) 13. (a) 14. (c) 15. (c) 16. (b) 17. (d) 18. (a) 19. (c) 20. (c) 21. (a) 22. (b) 23. (c) 24. (c) 25. (b) 26. (b) 27. (c) 28. (b) 29. (c) 30. (a) 31. (b) 32. (b) 33. (c) 34. (a) 35. (b) 36. (c) 37. (d) Subjective Questions 1. 5.59 nm 2. E1 = - 4613 eV, E3 = - 2650 eV 3. 0.48 mA 6. (a) 5.92 ´ 1020 Hz (b) 5.06 ´ 10–13 m 4. (a) 2.47 ´ 10–19 J = 1.54 eV (b) 804 nm, infrared 5. (a) 5.0 ´ 1014 Hz (b) 2.3 ´ 1020 photons/s 7. (a) K2 = 4K1 (b) E2 = 2E1 8. (a) 2.52 ´ 1019 (b) 3.33 ´ 10–8 N 9. 4.3 ´ 10–8 N 10. 10–8 N 11. 3.90 ´ 10–34 m, No 12. (a) 1.55 ´ 10–10 m (b) 8.44 ´ 10–14 m 13. (a) 2.37 ´ 10–24 kg- m (b) 3.07 ´ 10–18 J = 19.2 eV s 14. 1.04 Å 15. 6.663 Å 16. (a) 3.32 ´ 10–10 m (b) 1.33 ´ 10–9 m 17. 79 eV 18. 3, 6513 Å 19. 113.74 Å 20. n = 5 21. (a) Z = 4 (b) lmin = 40441 Å 22. (a) 15.6 volt (b) 2335 Å (c) 12.52 V (d) 1.01 ´ 107 m–1 23. (a) –5.08 eV (b) –5.63 eV

Chapter 33 Modern Physics - I — 323 24. 1.9 V 26. 27 l0 27. 49.5 pm, 99.0 pm 28. Z » 41 29. 15865 V 32 30. 198 pm 31. Z = 42 32. (a) 4 W (b) 396 J/s 33. 1022 Å, n = 3 to n = 1 34. 2.51 eV 35. 2260 Å 36. (a) 1015 Hz (b) 4 eV (c) 6.4 ´ 10–34 J-s 37. 1.81 eV, 9.0 ´ 105 m/s, 3.0 ´ 105 m/s 38. 0.148 m 39. 3.27 eV 40. 1.2 V LEVEL 2 Single Correct Option 1. (b) 2. (b) 3. (a) 4. (c) 5. (c) 6. (b) 7. (d) 8. (a) 9. (d) 10. (c) 11. (c) 12. (c) 13. (d) 14. (c) 15. (a) 16. (c) 17. (d) 18. (b) 19. (d) 20. (d) 21. (a) 22. (c) More than One Correct Options 1.(b,c) 2.(a,c) 3.(a,b,c) 4.(a,c) 5. (a,b,d) 6. (a,b) Comprehension Based Questions 1. (c) 2. (c) 3. (c) 4. (b) 5. (b) 6. (c) Match the Columns 1. (a) ® r (b) ® q (c) ® p (d) ® p (c) ® s (d) ® p 2. (a) ® s (b) ® r (c) ® r (d) ® s (c) ® s (d) ® q 3. (a) ® q (b) ® p (c) ® q (d) ® s (c) ® q (d) ® q 4. (a) ® r (b) ® s (c) ® q,s (d) ® p,r 5. (a) ® p (b) ® s 6. (a) ® r (b) ® p 7. (a) ® p,r (b) ® s Subjective Questions 1. (a) 1.31 mm (b) 164 nm 2. (a) 6.58 ´ 1015 Hz, 0.823 ´ 1015 Hz (b) 2.46 ´ 1015 Hz (c) 8.23 ´ 106 revolutions 3. (a) 2.55 ´ 10-13 m (b) 2.81 keV 4. (a) 102 nm, 122 nm, 651 nm (b) No lines 5. (a) 2.27 eV (b) 2.75 ´ 1017 (c) 9% 6. W = 11.925 eV 7. 2 eV, 0.754 V 8. neh , m 0pm2e7 9. 793.3 Å 10. 3.8 Å 4pm 8e0h5n5 11. (a) ni = 2 (b) Z = 6 (c) 28.43 Å (d) 489.6 eV, 25.3 Å 12. 4.5 eV 13. (a) 2000 Å, 1500 Å (b) E1 = - 8.25 eV,E2 = - 2.05 eV and E3 = - 0.95 eV (c) 8.25 V 14. (a) 4125 Å (b) 34 mA (c) 5.1% 15. 0.55 eV 16. 150 eV, 0.5 Å 17. (a) 1.69 ´ 10–28 kg (b) –2.53 keV (c) 0.653 nm 18. rn = nh , En = k ln íîì nh þýü 2p mk 2p mk 19. (a) rn = nh (b) K = nhBe (c) U = nheB (d) E = nheB (e) nh 2pBe 4pm 4pm 2pm 2e 20. (a) 10.2 volt (b) Z = 2 (c) 10.2 eV and 51 eV 21. 1.9 eV, 4125 Å, No change is observed 22. 1.6mA 23. (a) 0.31 eV (b) 4



34.1 Nuclear Stability and Radioactivity 34.2 The radioactive decay law 34.3 Successive disintegration 34.4 Equivalence of mass and energy 34.5 Binding energy and nuclear stability 34.6 Nuclear fission (Divide and conquer) 34.7 Nuclear fusion

326 — Optics and Modern Physics 34.1 Nuclear Stability and Radioactivity Among about 1500 known nuclides, less than 260 are stable. The others are unstable that decay to form other nuclides by emitting a and b-particles and g-electromagnetic waves. This process is called radioactivity. It was discovered in 1896 by Henry Becquerel. Whilst the chemical properties of an atom are governed N entirely by the number of protons in the nucleus (i.e. the 130 proton number Z), the stability of an atom appears to Number of neutrons 120 Line of depend on both the number of protons and the number of 110 stability neutrons. For light nuclei, the greatest stability is achieved 100 when the numbers of protons and neutrons are 90 N=Z approximately equal ( N » Z). 80 70 Z For heavier nuclei, instability caused by electrostatic 60 10 20 30 40 50 60 70 80 90 repulsion between the protons is minimized when there are 50 more neutrons than protons. 40 30 Figure shows a plot of N versus Z for the stable nuclei. For 20 mass numbers upto about A = 40, we see that N » Z. 40Ca is 10 the heaviest stable nucleus for which N = Z. For larger 0 values of Z, the (short range) nuclear force is unable to hold the nucleus together against the (long range) electrical Number of protons repulsion of the protons unless the number of neutrons Fig. 34.1 The stable nuclides plotted on a graph of neutron number, N, versus exceeds the number of protons. At Bi (Z = 83, A = 209), the proton number, Z. Note that for heavier nuclides, N is larger relative to Z. The neutron excess is N – Z = 43. There are no stable nuclides stable nuclides group along a curve with Z > 83. called the line of stability. The nuclide 209 Bi is the heaviest stable nucleus. 83 Atoms are radioactive if their nuclei are unstable and spontaneously (and randomly) emit various particles, the a, b and or g radiations. When naturally occurring nuclei are unstable, we call the phenomena natural radioactivity. Other nuclei can be transformed into radioactive nuclei by various means, typically involving irradiation by neutrons, this is called artificial radioactivity. A radioactive nucleus is called a parent nucleus, the nucleus resulting from its decay by particle emission is called daughter nucleus. Daughter nuclei also might be granddaughter nuclei, and so on. There are no son or grandson nuclei. For unstable nuclides and radioactivity, the following points can be made. (i) Disintegrations tend to produce new nuclides near the stability line and continue until a stable nuclide is formed. (ii) Radioactivity is a nuclear property, i.e. a, b and g emission take place from the nucleus. (iii) Nuclear processes involve huge amount of energy so the particle emission rate is independent of temperature and pressure. The rate depends solely on the concentration of the number of atoms of the radioactive substance. (iv) A radioactive substance is either an a-emitter or a b-emitter, g-rays emit with both.

Chapter 34 Modern Physics - II — 327 Alpha Decay An alpha particle is a helium nucleus. Thus, a nucleus emitting an alpha particle loses two protons and two neutrons. Therefore, the atomic number Z decreases by 2, the mass number A decreases by 4 and the neutron number N decreases by 2. The decay can be written as A X = A --24Y + 4 He Z Z 2 where, X is the parent nucleus and Y the daughter nucleus. As examples U 238 and Ra 226 are both alpha emitters and decay according to 238 U ¾® 23940Th + 4 He 92 2 226 Ra ¾® 222 Rn + 4 He 88 86 2 As a general rule, in any decay sum of mass numbers A 140 238U and atomic numbers Z must be the same on both sides. 234Th 234Pa Note that a nuclide below the stability line in Fig. 34.2 234U disintegrates in such a way that its proton number decreases and its neutron to proton ratio increases. In 230Th heavy nuclides, this can occur by alpha emission. 226Ra If the original nucleus has a mass number A that is 4 N=A–Z 222Rn times an integer, the daughter nucleus and all those in 214Pb 218Po the chain will also have mass numbers equal to 4 times an integer. (Because in a-decay A decreases by 4 and in 130 210Ti 214Bi b-decay it remains the same). Similarly, if the mass 210Pb 214Po number of the original nucleus is 4n +1, where n is an 210Bi integer, all the nuclei in the decay chain will have mass 210Po numbers given by 4n +1with n decreasing by 1 in each a-decay. We can see therefore, that there are four 206Pb a-decay possible a-decay chains, depending on whether A b-decay equals 4n, 4n +1, 4n + 2 or 4n + 3, where n is an integer. 80 84 88 92 Series 4n +1is now not found. Because its longest lived Z member (other than the stable end product Bi209) is Np237 which has a half-life of only 2 ´106 years. Fig. 34.2 The uranium decay series (A = 4n + 2). The decay of 21843Bi may proceed either by alpha emission and then beta Because this is much less than the age of the earth, this emission or in the reverse order. series has disappeared. Figure shows the uranium (4n + 2) series. The series branches at Bi214, which decays either by a-decay to Ti210 or b-decay to Po214. The branches meet at the lead isotope Pb210. Table 34.1 lists the four radioactive series. Table 34.1 Four Radioactive Series Mass numbers Series Parent Half-life, Years Stable product 4n Thorium 1.39 ´ 1010 Neptunium 232 Th 2.25 ´ 106 208 Pb 4n + 1 Uranium 90 4.47 ´ 109 82 4n + 2 Actinium 7.07 ´ 108 4n + 3 237 Np 209 Bi 93 83 238 U 206 Pb 92 82 235 U 207 Pb 92 82

328 — Optics and Modern Physics Beta Decay Beta decay can involve the emission of either electrons or positrons. A positron is a form of antimatter which has a charge equal to + e and mass equal to that of an electron. The electrons or positrons emitted in b-decay do not exist inside the nucleus. They are only created at the time of emission, just as photons are created when an atom makes a transition from a higher to a lower energy state. In b– decay a neutron in the nucleus is transformed into a proton, an electron and an antineutrino. n ¾® p + e– + n To conserve energy and momentum in the process, the emission of an antineutrino (n) (alongwith proton and electron) was first suggested by W. Pauli in 1930, but it was first observed experimentally in 1957. Thus, a parent nucleus with atomic number Z and mass number A decays by b– emission into a daughter with atomic number Z +1 and the same mass number A. A X ¾¾b– ® Z A Y Z +1 b– decay occurs in nuclei that have too many neutrons. An example of b– decay is the decay of carbon-14 into nitrogen, 14 C ¾® 14 N + e– +n 6 7 In b+ decay, a proton changes into a neutron with the emission of a positron (and a neutrino) p ¾® n + e+ + n Positron (e+ ) emission from a nucleus decreases the atomic number Z by 1 while keeping the same mass number A. A X ¾b¾+ ® Z -A1Y Z B + decay occurs in nuclei that have too few neutrons. A typical b+ decay is 13 N ¾® 163C + e+ + n 7 Electron capture Electron capture is competitive with positron emission since both processes lead to the same nuclear transformation. This occurs when a parent nucleus captures one of its own orbital electrons and emits a neutrino. A X + e–1 ¾® Z -A1Y + n Z In most cases, it is a K-shell electron that is captured, and for this reason the process is referred to as K-capture. One example is the capture of an electron by 4 Be 7 7 Be + e– ¾® 7 Li + n 4 3 Gamma Decay Very often a nucleus that undergoes radioactive decay (a or b-decay) is left in an excited energy state (analogous to the excited states of the orbiting electrons, except that the energy levels associated with

Chapter 34 Modern Physics - II — 329 the nucleus have much larger energy differences than those involved with the atomic electrons). The typical half-life of an excited nuclear state is 10-10 s. The excited nucleus ( X * ) then undergoes to a lower energy state by emitting a high energy photon, called the g-ray photon. The following sequence of events represents a typical situation in which g-decay occurs. 12 B ¾® 162C* + e– + n 5 12 C* ¾® 162C + g 6 12 B 5 e– 13.4 MeV e– 12 6 C* g 4.4 MeV 12 C 6 Gamma decay Fig. 34.3 Figure shows decay of B12 nucleus, which undergoes b-decay to either of two levels of C12. It can either decay directly to the ground state of C12 by emitting a 13.4 MeV electron or undergo b-decay to an excited state of 12 C* followed by g-decay to the ground state. The later process results in the 6 emission of a 9.0 MeV electron and a 4.4 MeV photon. The various pathways by which a radioactive nucleus can undergo decay are summarized in Table 34.2. Note In both a and b-decay, the Z value of a nucleus changes and the nucleus of one element becomes the nucleus of a different element. In g-decay, the element does not change, the nucleus merely goes from an excited state to a less excited state. Table 34.2 Various Decay Pathways Alpha decay A X ¾® A - 4 Y + 4 He Beta decay (b– ) Z Z - 2 2 Beta decay (b+ ) Electron capture A X ¾® Z + A Y + e– + n Gamma decay Z 1 ZAX ¾® Z - A Y + e+ + n 1 AZX + e– ¾® Z -A1Y + n A X* ¾® A X + g Z Z Extra Points to Remember ˜ After emission of one alpha particle and two beta particles isotopes are produced. This is because after the emission of one alpha particle, atomic number decreases by 2. Further, after the emission of two beta particles atomic number increases by 2. So, finally atomic number remains unchanged. ˜ From beta emission mass number does not change. Therefore, isobars will be produced.

330 — Optics and Modern Physics V Example 34.1 Mass number of a nucleus X is A and atomic number is Z. Find mass number and atomic number of the new nucleus (say Y) after the emission of m-alpha particles and n-beta particles. Solution By the emission of one alpha particle, atomic number decreases by 2 and by the emission of one beta particle atomic number increases by 1. Therefore, the atomic number of nucleus Y is Z = Z - 2m + n Further, by the emission of one alpha particle mass number decreases by 4 and by the emission of beta particle, mass number does not change. Therefore, the mass number of Y is A¢ = A - 4m 34.2 Radioactive Decay Law Radioactive decay is a random process. Each decay is an independent event and one cannot tell when a particular nucleus will decay. When a particular nucleus decays, it is transformed into another nuclide, which may or may not be radioactive. When there is a very large number of nuclei in a sample, the rate of decay is proportional to the number of nuclei, N , that are present çæ – dN ÷ö µ N è dt ø or æç – dN ÷ö = lN è dt ø where, l is called the decay constant. This equation may be expressed in the form dN = – ldt and N integrated, dN t ò òN N0 N =–l 0 dt or ln ççèæ N ÷÷öø = – lt N0 where, N 0 is the initial number of parent nuclei at t = 0.The number that survives at time t is therefore, N = N 0e–lt …(i) This function is plotted in Fig. 34.4. N N0 0.5 N0 0.37 N0 t1/2 tav t Fig. 34.4

Chapter 34 Modern Physics - II — 331 Half-life The time required for the number of parent nuclei to fall to 50% is called half-life t1/2 and may be related to l as follows. 0.5N 0 = N 0e– lt1/ 2 We have lt1/2 = ln (2) = 0.693 \\ t1/ 2 = ln (2) = 0.693 …(ii) l l Mean life The average or mean life tav is the reciprocal of the decay constant. tav =1 …(iii) l The mean life is analogous to the time constant in the exponential decrease in the charge on a capacitor in an RC circuit. After a time equal to the mean life time, the number of radioactive nuclei decreases to 1 times or approximately 37% of their original values. e Activity of a Radioactive Substance The decay rate R of a radioactive substance is the number of decays per second. And as we have seen above – dN µ N or – dN = lN dt dt Thus, R = – dN or R µ N or dt R = lN or R = lN 0e–lt or R = R0e–lt …(iv) where, R0 = lN 0 is the activity of the radioactive substance at time t = 0. The activity versus time graph is shown in Fig. 34.5. R R0 0.5 R0 0.37 R0 t1/2 tav t Fig. 34.5 Thus, the number of nuclei and hence the activity of the radioactive substance also decreases exponentially with time. Units of activity The SI unit for the decay rate is the Becquerel (Bq), but the curie (Ci) and rutherford (rd) are often used in practice. 1 Bq =1decays/s, 1Ci = 3.7 ´1010 Bq and 1 rd =106 Bq

332 — Optics and Modern Physics Extra Points to Remember ˜ After n half-lives, èçæ 1 ø÷ö n 2 (a) number of nuclei left = N0 (b) fraction of nuclei left = æèç 1 ö÷øn and 2 (c) percentage of nuclei left = 100 æçè 1 ø÷ö2 2 ˜ Number of nuclei decayed after time t, Number of nuclei decayed N0 = N0 – N = N0 – N0e–lt = N0 (1 – e–lt ) The corresponding graph is as shown in Fig. 34.6. ˜ Probability of a nucleus for survival upto time t, Time P (survival) = N = N0e–lt = e–lt N0 N0 Fig. 34.6 The corresponding graph is shown in Fig. 34.7. P(Survival) 1 Time Fig. 34.7 P (disintegration) 1 ˜ Probability of a nucleus to disintegrate in time t is, P (disintegration) = 1 – P (survival) = 1 – e–lt The corresponding graph is as shown in Fig. 34.8. ˜ Half-life and mean life are related to each other by the relation, t1/ 2 = 0.693 tav or t av = 1.44 t1/ 2 Time ˜ As we discussed above number of nuclei decayed in time t are Fig. 34.8 N0 (1 – e–lt ). This expression involves power of e. So, to avoid it we can use DN = lNDt where, DN are the number of nuclei decayed in time Dt at the instant when total number of nuclei are N. But, this can be applied only when Dt << t1/ 2. Proof - dN = lN Þ -d N = lNdt dt or DN = lNDt ˜ In same interval of time, equal percentage (or fraction) of nuclei are decayed (or left undecayed).

Chapter 34 Modern Physics - II — 333 V Example 34.2 At time t = 0, number of nuclei of a radioactive substance are 100. At t = 1 s these numbers become 90. Find the number of nuclei at t = 2 s. Solution In 1 second, 90% of the nuclei have remained undecayed, so in another 1 second 90% of 90, i.e. 81 nuclei will remain undecayed. V Example 34.3 At time t = 0, activity of a radioactive substance is 1600 Bq, at t = 8 s activity remains 100 Bq. Find the activity at t = 2 s. Solution R = R0 æç 1 ö÷ n è 2 ø Here, n is the number of half-lives. R = R0 Given, 16 R0 æç 1 ö÷ n 16 è 2 ø \\ = R0 or n=4 Four half-lives are equivalent to 8 s. Hence, 2 s is equal to one half-life. So, in one half-life activity will remain half of 1600 Bq, i.e. 800 Bq. V Example 34.4 From a radioactive substance n1 nuclei decay per second at an instant when total number of nuclei are n2 . Find half-life of the radioactive substance. Solution Using the equation, - dN - = lN dt We have, n1 = ln2 \\ l = n1 n2 Now, half-life is given by t1/ 2 = ln 2 = ln 2 ) l (n1 / n2 = çæçè n2 ÷ø÷ö ln 2 Ans. n1 V Example 34.5 Half-life of a radioactive substance is T. At time t1 activity of a radioactive substance is R1 and at time t2 it is R2 . Find the number of nuclei decayed in this interval of time. Solution Half-life is given by t1/ 2 = ln 2 l \\ l = ln 2 = ln 2 t1/2 T

334 — Optics and Modern Physics Activity R = lN \\ N = R = RT l ln 2 When activity is R1 , numbers of nuclei are N1 = R1T ln 2 Similarly, N2 = R2T ln 2 \\ Numbers decayed - R2 = N1 -N2 = (R1 ln 2 )T Ans. V 1E0xlamanpdlel3, 4re.6speTctwivoelrya.dIifoiancittiivaellymtahteeyrihaalsveXt1heansadmXe2nhuamvbeedreocfanyuccolneis,tathnetns the ratio of the number of nuclei of X 1 to that of X 2 will be 1/e after a time (JEE 2000) (a) 1 /10 l (b) 1 /11 l (c) 11 /10 l (d) 1 / 9 l Solution N x1 (t ) = 1 or N x2 (t ) e or N 0 e -10lt =1 (Initially, both have same number of nuclei say N 0 ) N 0 e-lt e e-9 lt = e-1 or e = e9lt or 9l t = 1 or t = 1 9l \\ The correct option is (d). V Example 34.7 The half-life period of a radioactive element x is same as the mean life time of another radioactive element y. Initially, both of them have the same number of atoms. Then, (JEE 1999) (a) x and y have the same decay rate initially (b) x and y decay at the same rate always (c) y will decay at a faster rate than x (d) x will decay at a faster rate than y Solution (t1/ 2 )x = (t mean ) y 0.693 = 1 or lx ly \\ lx = 0.693 l y or Rate of decay = lN lx < ly Initially, number of atoms (N) of both are equal but since l y > lx , therefore, y will decay at a faster rate than x. \\ The correct option is (c).

Chapter 34 Modern Physics - II — 335 V Example 34.8 A radioactive sample consists of two distinct species having equal number of atoms initially. The mean life of one species is t and that of the other is 5 t . The decay products in both cases are stable. A plot is made of the total number of radioactive nuclei as a function of time. Which of the following figure best represents the form of this plot? (JEE 2001) NN (a) (b) t tt t N N (d) (c) tt tt Fig. 34.9 Solution The total number of atoms can neither remain constant (as in option a) nor can ever increase (as in options b and c). They will continuously decrease with time. Therefore, (d) is the appropriate option. INTRODUCTORY EXERCISE 34.1 1. The decay constant of a radioactive sample is l. The half-life and mean life of the sample are respectively given by (JEE 1989) (a) 1/ l and (ln 2) / l (c) l (ln 2) and 1/ l (b) (ln 2) / l and 1/ l (d) l / (ln 2) and 1/ l 2. Consider a-particles, b-particles and g-rays each having an energy of 0.5 MeV. In increasing order of penetrating powers, the radiations are (JEE 1994) (a) a, b, g (b) a, g, b (c) b, g, a (d) g, b, a 3. Which of the following is a correct statement? (JEE 1999) (a) Beta rays are same as cathode rays (b) Gamma rays are high energy neutrons (c) Alpha particles are singly ionized helium atoms (d) Protons and neutrons have exactly the same mass 4. The electron emitted in beta radiation originates from (JEE 2001) (a) inner orbits of atom (b) free electrons existing in nuclei (c) decay of a neutron in a nucleus (d) photon escaping from the nucleus

336 — Optics and Modern Physics 5. During a negative beta decay, (JEE 1987) (a) an atomic electron is ejected (b) an electron which is already present within the nucleus is ejected (c) a neutron in the nucleus decays emitting an electron (d) a part of the binding energy of the nucleus is converted into an electron 6. A freshly prepared radioactive source of half-life 2 h emits radiation of intensity which is 64 times the permissible safe level. The minimum time after which it would be possible to work safely with this source is (JEE 1988) (a) 6 h (b) 12 h (c) 24 h (d) 128 h 7. A radioactive sample S1 having an activity of 5 mCi has twice the number of nuclei as another sample S2 which has an activity of 10 mCi. The half-lives of S1 and S2 can be (JEE 2008) (a) 20 yr and 5 yr, respectively (b) 20 yr and 10 yr, respectively (c) 10 yr each (d) 5 yr each 8. Half-life of a radioactive substance A is 4 days. The probability that a nucleus will decay in two half-lives is (JEE 2006) (a) 1 (b) 3 (c) 1 (d) 1 4 4 2 9. After 280 days, the activity of a radioactive sample is 6000 dps. The activity reduces to 3000 dps after another 140 days. The initial activity of the sample in dps is (JEE 2004) (a) 6000 (b) 9000 (c) 3000 (d) 24000 10. The half-life of 215At is 100 ms. The time taken for the activity of a sample of 215At to decay to 1 th of its initial value is (JEE 2002) 16 (a) 400 ms (b) 63 ms (c) 40 ms (d) 300 ms 11. The half-life of the radioactive radon is 3.8 days. The time, at the end of which 1/20 th of the radon sample will remain undecayed, is (given log10 e = 0.4343) (JEE 1981) (a) 3.8 days (b) 16.5 days (c) 33 days (d) 76 days 12. Activity of a radioactive substance decreases from 8000 Bq to 1000 Bq in 9 days. What is the half-life and average life of the radioactive substance? 13. A radioactive substance has a half-life of 64.8 h. A sample containing this isotope has an initial activity (t = 0) of 40 mCi. Calculate the number of nuclei that decay in the time interval between t1 = 10.0 h and t2 = 12.0 h. 14. A freshly prepared sample of a certain radioactive isotope has an activity of 10 mCi. After 4.0 h its activity is 8.00 mCi. (a) Find the decay constant and half-life (b) How many atoms of the isotope were contained in the freshly prepared sample? (c) What is the sample's activity 30.0 h after it is prepared? 15. A radioactive substance contains 1015 atoms and has an activity of 6.0 ´ 1011 Bq. What is its half-life? 16. Two radioactive elements X and Y have half-life periods of 50 minutes and 100 minutes, respectively. Initially, both of them contain equal number of atoms. Find the ratio of atoms left NX /NY after 200 minutes.

Chapter 34 Modern Physics - II — 337 34.3 Successive Disintegration Suppose a parent radioactive nucleus A (decay constant c=olnsat)ahnat s=nlubm)bwerhiocfhaitsomfusrtNhe0r at time t = 0. nucleus B (decay radioactive. After disintegration it converts into a Initially (t = 0), number of atoms of B are zero. We are interested in finding N b , the number of atoms of B at time t. laNa lbNb AB Fig. 34.10 AB At t = 0 N0 0 At t = t N a = N 0e–lat Nb =? At time t, net rate of formation of B = rate of disintegration of A – rate of disintegration of B \\ dN b =la Na – lb Nb dt or dN b = l a N 0e–lat – lb Nb (as N a = N 0e–lat ) dt or dN b + l b N b dt = l a N 0e–lat Multiplying this equation by elb t , we have elbt dN b + elbt l b N b dt = l a N 0e(lb - la ) t \\ d {N b elbt } = l a N 0e(lb - la ) t dt Integrating both sides, we get N b elbt = ççèæ l la ÷÷öø N 0e(lb – la ) t + C …(i) b –l a where, C is the constant of integration, which can be found as under. At time, t = 0, N b = 0 C = – æèçç l la ÷÷öø N 0 \\ b –l a Substituting this value in Eq. (i), we have Nb = N 0l a (e–lat – e–lbt ) …(ii) lb –la Now, the following conclusions may be drawn from the above discussion. 1. From Eq. (ii) we can see that N b = 0 at time t = 0 (it was given) and at t = ¥ (because B is also radioactive)