Tiêu đề 24. Atomic Physics 2 - Hard Ans.pdf ✅

1. (d) Using 2 r  n  2 1 2 1 2         = n n r r or 2 1 2 1 2         = n n d d  2 2 1 10 1.06       = d  d = 106 Å 2. (a) E 1.9 eV 36 5 13.6 3 1 2 1 13.6 2 2  =  =       = − 3. (d) Rydberg constant 2 2 2 0 mZe n h R   = Velocity nh Ze v 0 2 2 = and energy 2 2 2 0 2 4 8 n h mZ e E  = − Now, it is clear from above expressions R.v  n 4. (b) Energy difference between n = 2 and n = 3; E K K K 36 5 9 1 4 1 3 1 2 1 2 2  =       = −      = − .....(i) Ionization energy of hydrogen atom 1 n1 = and n2 =  ; E K  = K        = − 2 2 1 1 1 .....(ii) From equation (i) and (ii) E E 7.2 E 5 36  = = 5. (c) According to Bohr model time period of electron 3 T  n  1 8 1 2 3 3 3 1 3 2 1 2 = = = n n T T  T2 = 8T1 . 6. (d) Energy of a electron in nth orbit of a hydrogen like atom is given by eV n Z En 2 2 = −13.6 , and Z = 3 for Li Required energy for said transition        =        = − = − 9 8 13.6 3 3 1 1 1 E E E 13.6Z 2 2 2 2 3 1 108.8eV 108.8 1.6 10 J −19 = =   Now using  hc E =  E hc   =  m 7 19 34 8 0.11374 10 108 .8 1.6 10 6.6 10 3 10 − − − =        =   = 113 .74 Å 7. (d) For transition 3 → 1  hc E = 2E − E =   hc E = .....(i) For transition 2 → 1  − = hc E E 3 4   = hc E 3 .....(ii) From equation (i) and (ii)  = 3 8. (d) In the transition from orbits 5 → 2 more energy will be liberated as compared to transition from 4 → 2. So emitted photon would be of violet light. 9. (c) Using         = − 2 2 2 1 1 2 1 1 n n RZ         =   −  − 2 2 7 2 9 5 1 2 1 1.1 10 108 .5 10 1 Z  100 21 1.1 10 108 .5 10 1 7 2 9 =     − Z  4 108 .5 10 1.1 10 21 100 9 7 2 =     = − − Z  Z = 2 Now Energy in ground state E 13.6Z eV 13.6 2 eV 54.4 eV 2 2 = − = −  = − 10. (c) Using         = − 2 2 2 1 1 1 1 n n R         =  −  − 2 2 7 10 1 1 1 1.097 10 975 10 1 n  n = 4 Now number of spectral lines 6 2 4(4 1) 2 ( 1) = − = − = n n N . 11. (c) Let electron absorbing the photon energy reaches to the excited state n. Then using energy conservation  13.6 12.4 13.6 2 − = − + n  1.2 13.6 2 − = − n  12 1.2 2 13.6 n = =  n = 3.46 ≃ 4 12. (d) Time period 2 3 Z n T  For a given atom (Z = constant) So 3 T  n .....(i) and radius 2 R  n .....(ii)  From equation (i) and (ii) 3 / 2 T  R  8 1 4 3 / 2 3 / 2 2 1 2 1  =      =         = R R R R T T 13. (c) By using t n E t W P  = = where n = Number of uranium atom fissioned and E = Energy released due to each fission so 3600 170 10 1.6 10 300 10 6 19 6 −      = n  n = 40  1021
14. (c) Energy of an electron in ground state of an atom (Bohr’s hydrogen atom like) is given as E = - 13.6 Z2 eV where Z = atomic number of the atom  The ionization energy of that atom = Eion = 13.6 Z2  (Eion)H/Eion)Li = 2 Li 2 H (Z ) (Z )  9 1 3 1 (E ) (E ) 2 ion Li ion H  =      = 15. (c) Energy of radiation that corresponds to the energy difference between two energy levels n1 and n2 is given as E = 13.6         − 2 2 2 1 n 1 n 1 eV E is minimum when n1 = 1 & n2 = 2  Emin = 13.6 eV 4 3 eV 13.6 4 1 1 1  =       − E is maximum when n1 = 1 & n2 =  (the atom is ionized, that is known as ionization energy)  Emax = 13.6        − 1 1 = 13.6 eV.  4 3 E E max min =  4 3 hc / hc / min max =    4 3 max min =   16. (a) The energy of an electron in an orbit of principal quantum number n is given as E = eV n 13.6 2 −  -3.4 eV = eV n 13.6 2 −  n 2 = 4  n = 2 The angular momentum of an electron in nth orbit is given as L = 2 nh , Putting n = 2 We obtain L =  =  h 2 2h 17. (a) Frequency of revolution of electron is f  3 n 1 frequency of photon emitted           − − 2 2 n 1 (n 1) 1           − − − 2 2 2 n(n 1)] n (n 1)   2 2 n (n 1) [(2n 1)] − − when n >> 1   4 n 2n   3 n 1 18. (d) Energy of emitted photon is E =  hc = 2.54 ev The excitation energy is the energy to excite the atom to a level above the ground state. Therefore the energy level is E = – 13.6 + 10.19 = – 3.41 eV Photon arises from transition between energy state such that Ei – Ef = h = 2.54 eV Ei = 2.54 + Ef Ei = 2.54 + (E) = 2.54 – 3.41eV = – 0.87eV 19. (b) B = 2r μ i 0 ; magnetic field at centre of hydrogen atom i.e. at nucleus. i = T e = ef =  z 2 /n3 r  n 2 /Z B  i/r  Z 3 /n5 B  1/n5 20. (a) Shortest wavelength of Bracket series corresponds to the transition of electron n1 = 4 and n2 =  and the shortest wavelength of Balmer series corresponds to the transition of electron between n1 = 2 and n2 = . So (Z2 )       16 13.6 =       4 13.6  Z 2 = 4 or Z = 2 21. (a) E = E2 – E1 = 10.2eV = –3.4eV + 13.6eV so, n2 = 2 & n1 = 1 L = 2 3.14 6.63 10 2 h 2 h – 2 2h –34   =  =   J.s = 1.05 × 10–34J.s 22. (c) Maximum energy is liberated for transition En → 1 and minimum energy for En → En–1 Hence , 1 2 1 E n E − = 52.224 eV ... (1) 2 1 2 1 (n 1) E n E − − = 1.224 eV ... (2) Solving (1) and (2) E1 = – 54.4 eV E1 = 2 2 1 13.6Z − Z = 2 23. (a) Frequency of revolution of electron is
f  3 n 1 frequency of photon emitted           − − 2 2 n 1 (n 1) 1           − − − 2 2 2 [n(n 1)] n (n 1)   2 2 n (n 1) [(2n 1)] + − When n >> 1   4 n 2n   3 n 2 3 n 1   24. (a) The Bohr radius for a hydrogen-like atom of nuclear charge Z and one electron outside complete shells is me Z 2  = Zmc  . Thus the correct answer is (A) 25. (a) Shortest wavelength of Brackett series corresponds to the transition of electron between n1 = 4 and n2 =  and the shortest wavelength of Balmer series corresponds to the transition of electron between n1 = 2 and n2 = , So, (Z2 )       16 13.6 =       4 13.6  Z 2 = 4 or Z = 2 26. (b)  1 = R       − 2 n 1 1 1 2 n 1 = R 1 1  − n = R 1 R  −  27. (b) E = 13.6       − 2 2 6 1 1 1 = 13.22 eV p = c E  v = mC E = –27 8 –19 1.67 10 3 10 13.22 1.6 10      = 4.2 m/s 28. (b) Longest wavelength n = 2 to n = 1 r mv2 = 2 r k.Ze(2e) mvr = 2 nh Kinetic energy, potential energy and total energy becomes four times E becomes four times  E = 4Z2Rch       − 2 2 2 1 1 1 E = 3RchZ2  hc = 3RchZ2  = 3Z R 1 2 29. (d) v =       n Z 137 c  Z = 1, n = 1  c/137 30. (c) L =  − −  2 (n 1)h 2 nh =  +  −  2 h 2 nh 2 nh 31. (c) L = mvr = 2 nh L  n L  n  r r  n 2 r n 32. (b) In transition 2 → 1 maximum energy is emitted so frequency is maximum. 33. (d) A  r 2 r  n 2  A  n 4  1 2 A A = 1 16 34. (a) The Bohr radius for a hydrogen-like atom of nuclear charge Z and one electron outside complete shells is me Z 2  = Zmc  . Thus the correct answer is (A) 35. (a) For infrared  > 700 nm i.e., wavelength is greater than 700 nm
– 0.56 eV – 0.85 eV – 1.51 eV – 3.4 eV n =  n = 5 n = 4 n = 3 n = 2 n = 1 – 13.6 eV  1 =         − 2 i 2 f y n 1 n 1 R ni =  ; nf = 3  1 = 1.097 × 107       − 0 9 1  = 1.097 9 10−7   = 820 × 10–9 m 36. (a) Radius rn = 0.529 Z n 2 Å 37. (a)  1 = R         − 2 i 2 f n 1 n 1 nf = 2 ; ni = 3 R = 1.097 × 107 /m  = 656 nm 38. (c) Bohr has assumed stationary orbits in which there is no gain or loss of energy and angular momentum in any orbit does not change. 39. (b) r  n 2 –11 –11 21.2 10 5.3 10   = 2 n 1 n 2 = 4 , n = 2 40. (c) P = c E = mv P = c 13eV = mv 8 –27 –19 3 10 1.6 10 13 1.6 10      = v 41. (c) Due to 12.1 eV energy it is excited to 3rd orbit, so total lines = 2 n(n –1) = 2 3 2 = 3 lines 42. (a) PE = 2 × TE PE = 2 × ( – 13.6) = –27.2 eV 43. (d)  1 = R        − 1 1 = R   = R 1 44. (c) RN = me  RS = e e m m / 2 RS = 2 1 RS = 2 1 × 912Å 1 = 1824Å 1 45. (d) First excited state implies n = 2, while ground state implies n = 1. If radius of orbit in ground state, r1 = a0, then radius of orbit in first excited state = r2 = n2 r1 = (2)2 r1 = 4r1 = 4a0  A1 =  2 1 r =  2 0 a and A2 = 2 2 r = (4a0) 2 = 16 2 0 a Hence, (A2/A1) = (16/1) 46. (b) mvr = 2 nh ....(1) 2 2 r kZe = r mv2 ....(2) 47. (b) mvr = 2 nh , P = mv  n Z , Radius  Z n 2 energy  2 2 n Z 48. (c)  1 = RZ2         2 2 2 1 n 1 – n 1        =          =  1 – 1 1 R Z 1 1 – 1 1 R Z 912 1 2 2 2 2 49. (b)