Tiêu đề 24. Atomic Physics -1 Easy Ans.pdf ✅

1. (b) Electric field V m d V 10000 / 2.5 10 250 2 =  = = − . 2. (b) 3. (d) In Millikan's experiment, drops of non-volatile liquid (cloak oil) are used to prevent evaporation. 4. (b) E = eV = 2e  5 = 10 eV 5. (d) 19 5 = = 1.6 10 10 − E eV J 14 1.6 10 − =  6. (a) Any charge in the universe is given by n q q = ne  e = (where n is an integer) 1 2 3 4 5 6 1 2 3 4 5 6 q : q : q : q : q : q :: n : n : n : n : n : n 6.563 : 8.204 :11.5 :13.13 :16.48 :18.09 1 2 3 4 5 6 :: n : n : n : n : n : n Divide by 6.563 1 : 1.25 : 1.75 : 2.0 : 2.5 : 2.75 1 2 3 4 5 6 :: n : n : n : n : n : n Multiplied by 4 4 : 5 : 7 : 8 : 10 : 11 1 2 3 4 5 6 :: n : n : n : n : n : n 45 73.967 10 19 1 2 3 4 5 6 1 2 3 4 5 6 −  = + + + + + + + + + + = n n n n n n q q q q q q e C 19 1.641 10 − =  (Note : If you take 45.0743 in place of 45, you will get the exact value) 7. (d) Because magnetic force always points perpendicular to the particle velocity. That is why velocity remains unchanged thereby keeping energy       2 2 1 mv and momentum (mv) unchanged. 8. (b) 9. (c) Mass is basically a constant term for any physical application at low velocity. But in accordance with Einstein’s theory of relativity, at higher speeds the mass of the particle change according to formula 1 ( / ) 2 2 0 v c m m − = 10. (b) Here the velocity of electron increases, so as per Einstein’s equation mass of the electron increases, hence the specific charge m e decreases. 11. (c) If the voltage given is V, then the energy of electron m eV mv eV v 2 2 1 =  =  7 3 1 1 9 1.875 10 9.1 10 2 1.6 10 1000 =      = − − 1.9 10 m /s 7   12. (b) 13. (d) Momentum p = mv and m QV v 2 =  p = 2QmV  p  Qm    m m e m e m p pe e e 2 2 =   = 14. (b) In an electric field, a force opposite to the direction of electric field acts on negatively charged particles (i.e. from lower potential to higher potential). 15. (d) 16. (c) Ve mgd n E mg QE = mg  Q =  = 5 2 10 1.6 10 1.8 10 10 0.9 10 3 19 14 2 =         = − − − n 17. (c) 18. (b) In Millikan’s experiment, the charges present on the oil drops are the integral multiples, so 2e and 10 (1.6 10 ) 18 e C −  charges are present. 19. (c) m s B E eE evB v 1.5 10 / 2 10 3 10 7 3 4 =    =  = = − 20. (b) 21. (b) Charged particles trace a circular path in a perpendicular magnetic field. 22. (c) C k g m e 1.76 10 / 9.1 10 1.6 10 11 31 19 =    = − − 23. (c) 24. (d) Light consists of photons and cathode rays consists of electrons. However both effect the photographic plate. 25. (c) 26. (d) 27. (b) For ionisation, high energy electrons are required.
28. (b) 40 / sec. 0.5 20 m B E v = = = 29. (c) Higher the voltage, higher is the KE. Higher the work function, smaller is the KE. 30. (a) Time period of revolution of electron v r T   2 2 = = Hence corresponding electric current r ev T e i 2 = = 1 . 2 3.14 0.5 10 1.6 10 2 10 10 19 6 i = mA        = − − 31. (a) K Q V Joules 19 17 . 1.6 10 100 1.6 10 − − = =   =  32. (c) K = Q.V = 1e 1Volt = 1e V 33. (a) Kinetic energy  Potential difference 34. (a) In discharge tube cathode rays (a beam of negative particles) and canal rays (positive rays) moves opposite to each other. They will experience a magnetic force in the same direction, if a normal magnetic field is switched on 35. (b) 4 1.6 10 6.35 10 19 19    = = − − e Q n 36. (b) 37. (a) When cathode rays strike the metal plate, they transfer their energy to plate. 38. (d) Cathode rays are beam of electrons. 39. (b) K = QV =e V =eV 40. (a) V m e m QV mv QV v       =  = = 2 2 2 1 2 v 2 1.6 10 200 8 10 m / s 11 6  =    =  . 41. (c) Speed of the cathode rays is 10 m / sec 3 10 m / s 7 7 −  42. (d) d QV QE = mg  mg = 43. (d) 44. (c) In the condition of no deflection = 2  2 2VB E m e If m is increased by 208 times then B should be increased 208 = 14.4 times 45. (b) The colour of the positive column in a discharge tube depends on the type of gas e.g. For air, colour is purple red, for H2 , colour is Blue etc. 46. (c) 47. (a) m s m h m p v 7.25 10 / 9.1 10 10 6.6 10 6 31 10 34 =     = = = − − −  48. (d) Cathode rays are stream of negative charged particle, so they deflect in electric field. 49. (c) 3 3 2 4 2 2 2 2 2.5 10 (1.2 10 ) (3.6 10 ) 2 −      = = VB E m e 1.8 10 C / kg 11 =  . 50. (c) Specific charge m q = ; Ratio 2 1 =  =             =    m m q q m q m q p p p . 51. (c) B E v = ; where V m d V E 10 / 1 10 1000 5 2 =  = = −  v 10 m /s 1 10 5 5 = = . 52. (b) 53. (b) 54. (b) 55. (d) In Thomson’s mass spectrograph E|| B 56. (d) 57. (c) In the absence of electric field (i.e. E = 0) mg = 6rv ...(i) In the presence of Electric field mg + QE = 6r(2v) ...(ii)  B – F + F v mg D1 = 6rv mg D2 = 6r(2v) E D2 QE
When Electric field to reduced to E/2 mg + Q (E / 2) = 6r(v') ...(iii) After solving (i), (ii) and (iii) We get v v 2 3 ' = 58. (a) 3 1 1 9 9.1 10 2 2 1.6 10 45.5 − −     = = m eV v 4 10 m / s 6 =  59. (b) A t ne t Q i 1 4 1 9 6 1.8 10 1.6 10 28.8 10 − − = = =    =  = 29 A 60. (a)  me  mp  m                       m q m q m q e p 61. (b) Acceleration ( ) m e E m QE a 2 3 = = 62. (b) 2 200 (8.4 10 ) 2 2 1 2 6 2 2   =  = = V v m e mv eV 1.76 10 . 11 kg C =  63. (c) K Q. V (2e) 10 V 2 10 eV 2M eV 6 6 =  =  =  = 64. (c) Positive rays consist of positive ions. 65. (b) q m r qB mv r =  2  2 2  q m is maximum for + C 66. (b) m s B E v 3750 / 3 10 1.125 10 10 6 =   = = − − 67. (a) Positive rays was discovered by J.J. Thomson. 68. (a) 69. (d) If electron oscillate with a frequency of 1 GHz, it does not radiate any energy, which corresponds a definite wavelength. It only radiate when it jump from one orbit to another orbit. 70. (b) m eV v m eV eV mv v 2 2 2 1 2 2 =  =  = 71. (a) 72. (c) eE = mg  E mg e = 11 6 6 16 10 10 16 10 10 − − =    = C. 73. (b) 74. (c) According to de-Broglie hypothesis. 75. (a) mv h p h  = = 76. (a) : 2mE h mv h  = = 2 2 2m h  E =  is same for all, so m E 1  . Hence energy will be maximum for particle with lesser mass. 77. (a) Particle is photon and it travels with the velocity equal to light in vacuum. 78. (b) mE E h p h 1 ; 2  = =    (h and m = constant) 79. (a) 1 4 ; 1 2 2 1 1 1 2 2 = =  = = m m v v m v h m v h  80. (a) mE h mv h mv E mv mE 2 2 ; 2 1 2 =  =   = = 81. (d) Dual nature Diffraction Wave nature Photoelectric effect Particle nature       → → 82. (a) 2 nh mvr = According to Bohr’s theory   n mv h r n  =      2 = for n =1 ,  = 2r 83. (b) mE h 2  =  m 1   (E = same) 84. (a) 1 1 2 2 =    = = p p m m mE m h      85. (c) m Q V h mE h    2 2 = = On putting Q C 19 2 1.6 10 −  =   m mp k g 27 4 4 1.67 10 −  = =    Å V 0.101  = 86. (b) 87. (b) 1 2 2 1 1 2 E E mE E h =    =     2 1 1 2 1 0 1 0 4 0.5 10 10 E E E E =  =   − − Hence added energy = E2 − E1 = 3E1 D3 = 6r(2v) mg E/2 QE/2
88. (d) 3 1 1 9 3 4 2 9 10 80 1.6 10 6.6 10 2 − − −       = = mE h  = 1.4 Å 89. (c) mv m h 1  =    90. (b) If an electron and a photon propagates in the from of waves having the same wavelength, it implies that they have same momentum. This is according to de-Broglie equation,  1 p  91. (c) p p h 1  =    92. (d) In photoelectric effect particle nature of electron is shown. While in electron microscope, beam of electron is considered as electron wave. 93. (b) 2 particle 2 1 K = mv also mv h  =  2 . 2 1 2 particle vh v v h K  =       = ...(i)  hc Kphoton = ...(ii) 8 3 2 3 10 2.25 10 2 8 8 photon particle =     = = c v K K 94. (c) 3 2 3.14 5.13 10 10 2 2 11 9 =    =  = = − − r r n n     95. (a) By using m v h e  electron =  me e h v  = 7.25 10 / . 9.1 10 10 6.6 10 6 3 1 1 0 3 4 =  m s    = − − − 96. (a) By using mE h 2  = E = 10–32 J = Constant for both particles. Hence m 1   Since m p  me so .  p  e 97. (b) By using V 1    1 2 2 1 V V =    2 150 10 600 2 10 = = −   2 = 0.5 Å. 98. (b) mv rms h  =  0.66 Å 2 1.67 10 3 10 6.6 10 2 7 3 3 4 =      = − −  99. (c) p 1      = −  p p    =  p p  400 1 100 0 0.25 = = p p  p = 400 p0 . 100.(a) T neutron 1    1 2 2 1 T T =    2 300 1200 (273 27) (273 927 ) 2 = = + + =    . 2 2   = 101.(d) m E mE h 1 2  =   ( = constant)  me  m p so Ee  Ep 102. (b) 103.(a) Wavelength of photon will be greater than that of electron because mass of photon is less than that of electron   ph  e 104.(b) 2 2 2 2   m h E mE h =  = 31 9 2 34 2 2 9.1 10 (0.3 10 ) (6.6 10 ) − − −      = J 18 2.65 10 − =  = 16.8 eV 105.(c) mQV mQ h 1 2  =    p p p m Q mQ    = 2 2 4 2 =   = p p p p m Q m Q 106.(a) 6 34 2 10 6.63 10 − −   =  = =   h p p h 3.31 10 - / sec 28kg m − =  107.(a) m mv h 37 34 3.3 10 1 2000 6.6 10 − − =     = = Å 27 3.3 10 − =  108.(a) 3 1 1 9 3 4 2 9.1 10 5 1.6 10 6.6 10 2 − − −       = = mE h  5.469 10 m 5.47 Å 10 =  = − 109.(c) 2 9.1 10 1.6 10 100 6.6 10 2 3 1 1 9 3 4       = = − − − mQV h  = 1.23 Å 110.(c) The De-Broglie wavelength is | | | I| h p h  = = | | 1 I    111.(d) Davission and Germer proved the wave nature of electron by performing an experiment.