7. Solution-Modern Physics

SOLUTIONS TO PART TEST-01 TOPIC : MODERN PHYSICS 1. Rate of decay of A keeps on decreasing continuously because concentration of A decreases with time  A is false Initial rate of production of B is 1N0 and rate of decay is zero. With time, as the number of B atoms increase, the rate of its production decreases and its rate of decay increases. Thus the number of nuclei of B will first increase and then decrease  B is the correct choice The initial activity of B is zero whereas initial activity of A is 1N0  C is false. As time t   : NA = 0, NB = 0 and NC = N0  D is false 2. E = 3 kT 2 & P = de–Broglie = h  h p de–Broglie = Substituting values :  = 0.63 Å hc 3.  = 5 eV0 +  hc 3 = eV0 +   2 hc 3 = 4eV0   = hc 6 h 4. Change in momentum due to photon =  F = rate of change of momentum h F = n  = ma  a = nh m h 5.  = p =   For higher m and q ;  will be smaller. For an '' particle; both 'm' and 'q' are higher hence lesser is the wavelength. 1 As, (penetrating power)  Energy   From above; penetrating power of an -particle is more than that of a proton.  dN   dt  (N) 𝑙n2 .N0 e  𝑙n2.2 1 2 6.  A  A = 1 = 2  2 = 1  dN  (N)B 𝑙n2  𝑙n2.2 21  dt  .N0 e 2  B 2 7. Mass defect =(238.05079 – 234.04363 – 4.00260) u = 4.56 × 10–3 u = 4.56 × 10–3 × 1.66 × 10–27 = 7.57 × 10–30 kg mc2 = 7.57 × 10–30 × 9 × 1016 = 6.8 × 10–13 J 8. The correct statement in nuclear force is not a central force. 9. Cut- off wavelength of emitted X-rays depends on maximum kinetic energy of incident electrons on the target and is independent of nature of target. The characteristic lines depend on nature of material of target. 10. The maximum K.E. of ejected photoelectrons is (K.E)max = h – 0 If the frequency of photon is doubled, maximum kinetic energy of photon electrons becomes K.E. = 2h – 0 ; K.E. (K.E.)max 2(h  0 ) = 2 > 2 h   0 Photo current  intensity of beam h If intensity and frequency both are doubled, the photocurrent remains same. 11. Energy of nth state in Hydrogen is same as energy of 3nth state in Li++.  3  1 transition in H would give same energy as the 3×3  1× 3 transition in Li++. nh 12. Angular momentum = 2 h = 2 ( n = 1) h 13.  =  2  1 K = qV is same for both proton and electron. 14. In the first case K.E. of H-atom increases due to recoil whereas in the second case K.E. decreases due to recoil.  E > E 15. Linear momentum  mv  1 n angular momentum  mvr  n  product of linear momentum and angular momentum  n0 16. For 2nd line of Balmer seires in hydrogen specturm 1  R(1) 1  1  3   22 42  = 16 R   For Li2+ : which is satisfied by transition n = 12 to n = 6 17. For an electron with kinetic energy E :  = and for a photon :  = hc E  hc 2mE Ratio = E . h = 18. Doubly ionised positively lithium ion is a hydrogen like atom : nh hence : L = m v r = 2 for both which depends only on the value of n. Also for both n = 1. h Hence; L1 = L2 = 2 ; 19. Potential energy = 2 Total energy = 2 × – 21.76 × 10–19 = – 43.52 × 10–19 J 20. As the accelerating potential difference is changed only the minimum wavelength changes. It has no effect on wavelengths of characteristic x-rays (whether they are produced or not) 21. The electron ejected with maximum speed vmax are stopped by electric field E = 4N/C after travelling a distance d = 1m  1 mV 2 = eE d = 4eV 2 max 1240 The energy of incident photon = 200 From equation of photo electric effect = 6.2 eV 1 mv 2 = h –    = 6.2 – 4 = 2.2 eV. 2 max 0 0 22. The energy of K x-ray photons is directly propotional to (Z–1)2 . The energy ratio of two K photons obtained  Z  1 2 in x-ray from two metal targets of atomic numbers Z1 and Z2 is  1   Z2  1 23. eVS = hc    1240(nm)eV  1.9 eV 400(nm)  = 1.2 eV VS = 1.2 V  The cesium ball can be charged to a maximum potential of 1.2 V. 24. h = eV   = eV h 1.6 10 19  10 103 = 6.6 10 34 = 2.4 × 1018 Hz 25. Photon has rest mass zero 26. Energy released = (80 × 7 + 120 × 8 – 200 × 6.5) = 220 MeV. 27. Endproduct doesnot decay further h 28.  = p = h2  1 1   E = h2 2m2 E =  2 2  2m  1 2  Put  = 0.5 × 10–9 m &  = 2 × 10–9 m and solve. 29. A = A0 e–t after one year : A0 = A 10 0 e–  e– = 1 10 After further 9 years (i.e. after 10 years) : A = A0 e–10 = A0 1010 . 30. 3Li7 + 0n1  –1e0 + 22He4 Hence the unknown nuclide is 3Li7 . SOLUTIONS TO PART TEST-02 TOPIC : MODERN PHYSICS 1. Using Mosely’s law for both cobalt and impurity = K (Z – 1)  = K(Z – 1)  = K (Zco – 1) and = K (Zx – 1)  Zx  1  Zco  1 = Zx = 40 2. The hydrogen atom is in n = 5 state.  Max. no of possible photons = 4 To emit photon in ultra violet region, it must jump to n = 1, because only Lyman series lies in u.v. region. Once it jumps to n = 1 photon, it reaches to its ground state and no more photons can be emitted. So only one photon in u. v. range can be emitted. If H atom emits a photon and then another photon of Balmer series, option D will be correct. 3. The energy of incident photons is given by h = eV +  = 2 + 5 = 7 eV (Vs P is stopping potential and  105P is work function ) Saturation current = 10–5 A =  P = 7 W. h e = 7  e e ( is photo emission efficiency)  1  1  3 4. First excitation energy = RhC  12 22  = RhC   4  3 RhC = V e.v.  RhC = 4V 4 3 e.v. 1 5.   3R (Z  1)2 4  (Z – 1) = = = = 78 = 26. Z = 27 3 R  z 1/ 3  z  27 6. R  A1/3 and here A = 2 Z 1 1 =  1   R2  z2  r1 z2 8  1  =  z2  8 Radius of Bohr’s orbit r  z r2 = z1 = 27 1 7.  2 1N =  2N decay rate of  decay = decayrate of decay probability of  decay = probability of  decay 75 100  3 = 25 100 8. N = N e–t N = N (1 – e–t) Rate of formation of Y = dN = + N e–t dt X  Y At t = 0 , R = N t = 0 N 0 t =  , R = 0 t = t N N = N (1 – e–t) 9. Given,  = 0.173 ln 2 T1/2 =  Also 0.693 = 0.173  4 N0 – N = N0 e–t 1 for t = 0.173 year : N0 – N = N0 = 0.37 N e 0 10.  hc min = eV = 62.1 pm 11. In ground state n = 1 and for first excited state n = 2 KE = 1 40 e (z = 1) = 2r 14.4 1010 2r eV ( r = 0.53 n2 Aº (z = 1)) (KE)1 = 14.4 1010 2  0.53 1010 eV = 13.58 eV and (KE)2 = 14.4 10 10 2  0.53  1010  4 ev = 3.39 ev  KE decreases by = 10.2 ev  PE increases by = Excitation energy + Loss in kinetic energy = 10.2 + 10.2 = 20.4 ev nh Now Angular momentum ; L = mvr =  L –L = h 6.6  1034 = = 1.05 × 10–34 J–sec. 2 2 1 2 6.28 12. As 1  t1/ 2  t > t  𝑙n2 (1/2) A (1/2)B at t = 60 min.   N =  N  N > N Activity curves are (T1/2 (A) > T1/2 (B)) Activity B A 1h Activity of B decreases rapidly than that of A. t 13. Both statements I and II are true; but even if radiation of single wavelength is incident on photosensitive surface, electrons of different KE will be emitted. 14. Neutron will knock out electrons in target leading to production of characteristic X-ray. Hence statement-1 is false. hc 15. KE = eV = hmax = min min = hc  eV 12400 10 103 = 1.24 Å (x-rays) x-rays are highly penetrating radiation. 16. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 17. Q = CV  ne = 2.85  1012 10 0 A V d n = 0.5 103  1.6  10 19  16 n = 8.85 × 109 nhC 18. P =  where n = no. of photons incident per unit time. hC Also,  = ne  P = e (2 106 )(6.6 10 34 )(3 108 ) = (4  106 )(1.6  1019 ) = 9.9 10 7 m = 1.6 9900 1.6 Å = 6187 Å Which came in the range of orange light. 19.  Stopping potential VS = 8V and KE = eVS  KE = 8eV 20. Total energy released from Au198  Hg198 in ground state = (m ) c2 = (197.9682 – 197.9662) (930) = 1.86 MeV Energy released from 198Hg in first excited state  Hg in ground state = (–1.6) – (–2) MeV = 0.4 MeV  Energy released from Ag198  Hg198 second excited state = 1.86 – 0.4 = 1.46 MeV = max. K.E. of  particle 21. Similarly maximum kinetic energy of  particle = 1.86 – 1 = 0.86 MeV 22.      1 3 2 0i q e 0e v v (z / n)  z3 23. (a) B = 2r where ieq = T = 2r / v  B = 42 r 2  r 2  (n2 / z)2 n5  q  2 e 2  n2   z  (b) Magnetic moment M = i A =  (r ) = r  r v       T  2r / v  z   n  (c)  = h  1  n mv (z / n) z L   nh / 2 (d) Areal velocity = 2m = 2  n SOLUTIONS TO PART TEST-03 TOPIC : MODERN PHYSICS 1  1  1  1. Using  = R(z – 1)2 n 2 n 2   2 1  For  particle ; n = 2, n = 1 For metal A ; 1875R = RZ  12  3   z = 26 4 For metal B; 675R = RZ  4   12  3  1  z = 31  4  2 Therefore, 4 elements lie between A and B. 2. As proton mass is less than neutron mass proton does not decay to neutron outside nucleus. 3. Pi =  = 0 For completely inelastic collision both come to rest af ter collision and net energy of 4E + E = 10.5 eV is lost. But electron in ground state of H-atom can accept only an energy of 10.2 eV. Hence the collision may be inelastic but it can never be perfectly inelastic. 4. Maximum energy of radiation incident on H-sample = KEmax of electron + 13.6 eV = 51eV this energy corresponds to the transition n = 4  n = 1 in Helium For electrons of the He to get excited to n = 4 12400  = (51) = 243 Å 5. Potential of the sphere at any time V(t)  Q0  Qt 40R = V + Pet 40Rhc (because Qt  P n e  t ) hc 6. Loss in K.E. will be used in splitting of Deutron.  Let us calculate the minimum K.E. required for the loss of 2.2 MeV. For this we must assume perfectly inelastic collision. 4m.u 2u = 6m = 3 maximum loss of K.E. will be in perfectly inelastic collision. By energy conservation.  1 × 4 mu2 – 1 2 2  2u 2 × 6m  3  = Eb   1  2  2 × 4mu2 1 3  = Eb    1 × 4mu2 = 3E 2 b ∴ KE > 3Eb KE > 6.6 MeV 7. no of moles of H2 consumed 1MW  (24  3600) sec/ day = (20 MeV  6.023  1023 ) = 0.05  m = 0.1 g dN 8. dt = t2 – N dN for dt d2N to be minimum; dt2 d2N = 0  dt2 dN = 2t –  dt = 2t –  (t2 – N) = 0 or N = 2t0  t2 2 1 9. v  z and rn  z 10. Ionization energy = 13.6 Z2 eV  (Ionization energy) > (Ionization energy)  Z > Z As u  Z  u > u 11. K x-ray is emitted when electron jumps from outer shell to K shell. This is equivalent to hole jumping from K shell to outer shell. <  12. Magnetic field at centre (site of nucleus) B = 0  0 qf = 0qv ]  B 1 and B  v 2r 2r  B  1 n5 2r  2r r 2 B1  (2)5 B2 (1)5 (Since , n1 = 1 to n2 = 2)  B1 = 32 B2 h h Also, mvr = n. 2 ,therefore angular momentum is decreased by 2 . 13. Statement-2 is true by definition and correctly explains the staement-1, namely, Z XA undergoes 2  decays, 2 decays (negative ) and 2  decays. As a result the daughter product is Z2 YA8 . 14. As electron jumps from n = 2 to n = 1, angular momentum ( n ) does not remain conserv ed. Hence statement-1 is false. 15. Fz = z . dBz dz = (9.3 × 10–24 J/T). (16 × 10–3 T/m) = 1.5 × 10–25 N 16. Energy change is B × 2 as the spin is completely flipped. U = 2B = 2 × (5.8 × 10–5 eV/T) × (620 × 10–3 T) = 7.2 × 10–5 eV 17. h  E  h = E  = E/h (7.2 105 eV) = (4.4 1015 eV.s) = 1.74 × 1010 Hz. 18. (A) In the given spontaneous radioactive decay, the number of protons remain constant and all conservation principles are obeyed. (B) In fusion reaction of two hydrogen nuclei, a proton is decreased as positron shall be emitted in the reaction. All the three conservation principles are obeyed. (C) In the given fission reaction the number of protons remain constant and all conservation principles are obeyed. (D) In beta negative decay, a neutron is converted into a proton and the electron is ejected out. 19. In one half-life the number of active nuclei reduces to half the original number. Thus, in two half-lives the  1   1  number is reduced to  2   2  of the original number. The number of remaining active nuclei is, therefore,      1   1  12 × 1018 ×  2   2  = 3 × 1018 = 3 Ans.     20. The volume of liquid in beaker at any instant of time t is V = 100 + 8t The volume of liquid ejected in t seconds is 2t  Number of active atoms being taken out is –dN = N V dN 2dt or – dt = 2N V 2N = 100  8t multiplying both sides with disintegration constant. –  dN = N 2dt V or – dA = A. 2dt V where A is activity of the solution. The time taken for 10 ml solution to come out is 5 second. A dA 5  2t  5 1/ 4 or  A =  100  8t dt or A = A0  7  A 0 0     5 1/ 4   required activity of the ejected solution is A – A = A 1    where A0 = 50 dps 0 0   7   12400 21. Energy of incident photons = 2000 eV. 12400 maximum kinetic energy of ejected electrons = 2000 –  12400 maximum kinetic energy of electrons striking plate B = 2000 –  + 10 12400 minimum wavelength of photons emitted from B = 1000 12400 12400  –  + 10 =   = 3.8eV 2000 1000 22.  = hc , ' hC = given 4 ( –  ) = ( – ’ ) min 20000e min 10000e  min  min  3 = 4 min   = hC 30000e  hC = 30000 e  By using the formula of energy of electrons according to Bohr’s model and considering shielding effect  1 1  30000 = (13.6)  12  2  (z – 1)2  2   z – 1 = 100 = = 54  z = 55