Title 24. Atomic Physics - Medium-1 ans.pdf ✅

1. (c) Sir J.J. Thomson proposed the first model of atom called plum pudding model of atom. According to this model, the positive charge of the atom is uniformly distributed throughout the volume of the atom and the negatively charged electrons are embedded in it like seeds in a watermelon. 2. (a) 3. (c) 10 m to10 m −15 −14 4. (a) The number of scattered particles detected will be maximum at the angle of scattering  = 0o at minimum at  =180o 5. (b) At minimum impact parameter  particles rebound back (  ) and suffers large scattering. 6. (c) The radius of orbit of electron 10 m 10 = Radius of nucleus 10 m −15 =  Ratio 5 15 10 10 10 10 = = − − Hence the radius of electron orbit is 5 10 times larger than the radius of nucleus. 7. (c) In hydrogen atom electrostatic force of attraction (Fe) between the revolving electrons and the nucleus provides the requisite centripetal force (fc) to keep them in their orbits. Thus, Fe = Fc 2 2 0 2 r e 4 1 r mv   = Or 4 mr e v 4 mr e v 0 2 0 2 2   =  = 8. (c) Trajectory of −particle depends on impact parameter which is the perpendicular distance of the initial velocity vector of the  particle from the center of the nucleus for small impact parameter  particle close to the nucleus and suffers larger scattering. 9. (c) Led d be the distance of closest approach then by the conservation of energy Initial kinetic energy of incoming  particle, k = Final electric potential energy U of the system As d (2e)(Ze) 4 1 K 0 = K 2Ze 4 1 d 2 0  = ..... (i) Here , 9 10 Nm C ,Z 79,e 1.6 10 C 4 1 9 2 2 19 0 − − =  = =   K 7.7MeV 7.7 10 1.6 10 J 1.2 10 J 6 −19 −12 = =    =  substituting these values in (i) 12 9 19 2 1.2 10 2 9 10 (1.6 10 ) 79 d − −       = d 3 10 m −14 =  = 30m ( 1fm 10 m) −15  = 10. (b) Kinetic energy 8 r e mv 2 1 K 0 2 2  = = And P.E. 4 r e U 0 2  = − Or 8 r e mv 2 1 K 0 2 2  = = U = −2K 11. (d) 15 15 3 10 3 10 (10 ) 3 4 (10 ) 3 4 Volume of nucleus Volume of atom =   = − − 12. (c) A few some most 13. (a) Emission spectrum is observed 14. (a) In 1885 the first spectral series were observed by a Swedish School teacher Johann Jakob Balmer. This series is called the Balmer series. 15. (d) Wavelength of Balmer series is       = −  2 2 n 1 2 1 R 1 At n =  , the limit of the series observed        = −   2 1 4 1 R 1 R 4 or 4 1 R =  =  Here, Rydberg’s constant 7 1 R 1.097 10 m − =  7 1.097 10 4   = 364.6 10 m 364.6nm 9 =  = − 16. (b) Balmer series 17. (d)
Since out of the given four lines H line has smallest wavelength hence the frequency of this line will be maximum. 18. (d) The wavelength of the Paschen series       = −  2 2 n 1 3 1 R 1 For shortest wavelength n =  9 1 R 9 1 R 1 2 =       = −   7 1.097 10 9 R 9   = = 8.20 10 m 820nm. 7 =  = − 19. (c) The wavelength for Pfund series is given by       = −  2 2 n 1 5 1 R 1 For series limit n =  25 1 R 25 1 R 1 2 =       = −   2278nm. 1.097 10 25 R 25 7 =   = = 20. (d) 100 21R 25 1 4 1 R 5 1 2 1 R 1 2 2 =      = −       = −  10 m 23.1 100 21 1.1 10 100 21R 100 7 7 − =     = = = 4346Å This correspond to violet. 21. (c) Here L =1215Å For the first line of Lyman series 4 3R 4 1 R 1 2 1 1 1 R 1 2 2 L =      = −       = −  4 1 3R L =   For first line of Balmer series       =         = −       = −  36 5 R 1 9 1 4 1 R 3 1 2 1 R 1 B 2 2 B 5R 1 36 B =   From (i) and (ii) 4 5 36 3 4 / 3R 36/ 5R L B   = =   1215 6561Å 20 108 20 108 B =   L =  = 22. (b) Jump to second orbit leads to Balmer series when an electron jumps from 4th orbit to 2nd orbit shall give rise to second line of Balmer series 23. (b) If we measure the frequencies of light emitted due to transition between excited states and the first exited state. 24. (d) The wavelength of the first line of Hyman series for hydrogen atom is       = −  2 2 2 1 1 1 R 1 The wavelength of the second line of Balmer series for hydrogen like ion is       = −  2 2 2 4 1 2 1 Z R ' 1 According to the question  = '       = −        − 2 2 2 2 2 4 1 2 1 Z R 2 1 1 1 R Or 16 3Z 4 3 2 = or Z 4 or Z 2 2 = = 25. (d) By Bohr’s formula       = −  2 2 2 1 2 n 1 n 1 Z R 1 For first line of Lyman series n 1,n 2 1 = 2 = 4 3 Z R 1 2 =   In the case of hydrogen atom Z = 1 For hydrogen like atom Z = 11 121 1 121R 3 4 4 3R ; ' 4 3 121R ' 1 =  =    =   10Å 121 1210 121 ' = =   = 26. (d) By Bohr’s formula       = −  2 2 2 1 2 n 1 n 1 Z R 1 For first line of Lyman series n 1,n 2 1 = 2 = 4 3 Z R 1 2 =   In the case of hydrogen atom Z = 1 For hydrogen like atom Z = 11 121 1 121R 3 4 4 3R ; ' 4 3 121R ' 1 =  =    =   10Å 121 1210 121 ' = =   = 27. (a) For Lyman series
       = − 2 2 n 1 1 1 R Where n = 2, 3, 4.......... For the series limit of Lyman series n =  Rc 1 1 1 Rc 1 2 2 =        = − ...... (i) For the first line of Lyman series n = 2 Rc 4 3 2 1 1 1 Rc 2 2 2 =       = − ...... (ii) For Balmer series        = − 2 2 n 1 1 1 R Where n = 3, 4, 5.......... For the series limit of Balmer series n =  4 1 Rc 2 1 Rc 3 2 2 =        = − ...... (iii) From equation (i), (ii) and (iii) we get 1 = 2 + 3 or 1 − 2 = 3 28. (c) For Balmer series n 2,n 3 1 = 2 = for 1st line and n 4 2 = for second line. 10 27 5 36 16 3 5/ 36 3/16 3 1 2 1 4 1 2 1 2 2 2 2 2 1 = =  =       −       − =   6561 4860Å 27 20 27 20 2 = 1 =  = 29. (d) 4 →3 30. (a) Bohr made a hypothesis that there are certain special state of motion called stationary state in which the electron may exist without radiating electromagnetic energy in these states according to Bohr, the angular momentum of electrons takes values that are integer multiples of h. in stationary states, the angular momentum of the electron may have magnitude h, 2h,3h..... but never such as 2.5 h or 3.1 h. this is called the quantization of angular momentum. 31. (c) The kinetic energy of the electron in hydrogen atom are        =  = = 4 mr e v 8 r e mv 2 1 K 0 2 2 0 2 2  Electrostatic potential energy 4 r e U 0 2  − = The total energy E of the electron in a hydrogen atom is E = K + U 8 r e 4 r e 8 r e E 0 2 0 2 0 2  = −          − +  = Here negative sign shows that electron is bound to the nucleus. 32. (a) Radius of first orbit, , Z 1 r  For doubly ionized lithium Z (=3) will be maximum hence for doubly ionized lithium r will be minimum 33. (a) According to Bohr’ second postulate Angular momentum  = 2 nh L Angular momentum is also called a moment of momentum For second orbit, n =2  =  = h 2 2h L 34. (b) Angular momentum nh 2 nh L mv r n n n =  = = Where h = planks constant rn = Radius of nth orbit n v = Velocity of electron in nth orbit. 35. (b) Angular momentum  = = 2 nh L mvr or 2 r nh mv  = Now, 2 r  n 2 2 n nh mv p   =  n 1 or p 2 n h p    Energy 2 n 1 E  36. (b) eV n 13.6 2 − 37. (c) The radius of nth orbit 2 0 2 2 n me h 4 r n  = Where 2 0 0 2 a me h 4 =  (Bohr radius) Hence, 0 2 n r = n a 38. (c) In hydrogen atom the lowest orbit corresponds to minimum energy 39. (d) The minimum energy required to free the electron from the ground state of the hydrogen atom is called the ionization energy of hydrogen atom its value is 13.6 eV. 40. (a) Absorption is from the ground state n = 1 to n’ where n' 1
41. (a) In a hydrogen atom the energy of electron in nth orbit is eV n 13.6 En 2 = − 42. (a) Here, E 13.6eV 13.6 1.6 10 2.2 10 J −19 −18 = − =   =  8 E e E 0 2  =  As orbital radius 8 E e r 0 2  − = 2 ( 2.2 10 ) 9 10 (1.6 10 ) 18 9 19 2 − −  −     = 5.3 10 m −11 =  43. (d) The first three options (a), (b) and (c) are Bohr’s postulates of atomic model whereas option (d) is not correct as Bohr’s model is applicable to hydrogen atom only. 44. (c) As 0 2 n r = n a Here a 0 = 0.529Å and n = 3 0 2 n 3 r = (3) a = = 90.529Å = 4.761Å 45. (b) Energy, eV n 13.6 En 2 = − In ground state energy 13.6eV 1 13.6 E1 2 = = − In first excited state energy 3.4eV 2 13.6 E2 2 = = − Then the required energy = E2 − E1 = −3.4 − (−13.6) =10.2 eV 46. (b) Radius of nth orbit in hydrogen like atom is Z a n r 2 0 n = Where 0 a is the Bohr’s radius For hydrogen atom Z = 1 1 0 r = a (  n = 1 for ground state) For Be ,Z 4 3 = + 4 a n r 2 0  n = According to given problem 1 n r = r 4 n a a 0 2 0 = n = 2 47. (d) As, P.E. = -2K.E. Here, K.E. 13.6eV 2.18 10 J −18 = =  Hence, P.E. 2 2.18 10 J −18 = −   4.36 10 J −18 = −  48. (b) As 0 2 n r = n a Here a 5.3 10 m 11 0 − =  n = 2 0 2 2  r = (2) a 4a 4 5.3 10 m 11 0 − = =   21.2 10 m 2.12Å 11 =  = − 49. (c) Here n1 =1 and n 4 2 = Energy of photon absorbed, E = E2 − E1 Since, eV n 13.6 En 2 = − Then         − = − − − 2 1 2 2 (1) 13.6 (4) 13.6 E E eV 16 13.6 15 13.6 16 13.6  = + = 12.75eV 12.75 1.6 10 J −19 = =   20.4 10 J −18 =   − = hc E2 E1 18 34 8 2 1 20.4 10 6.6 10 3 10 E E hc − −     = −  = 9 70 10 m 970 10 970Å 8 10 = −  =  = − − 50. (a) Since   = c Here 8 1 c 3 10 ms− =  9.7 10 m −8  =  3.1 10 Hz 9.7 10 3. 10 15 8 8 =   −   = 51. (c) According to Bohr’s model nh 2Ke Z v 2 = n 1 v  A B B A n n V V  = Here, 6 1 VA 2.2 10 ms− =  n A = 1,n B = 4 n B A B A n n V = v  6 6 5 1 0.55 10 5.5 10 ms 4 1 2.2 10 − =   =  = 