Irodov – Problems in General Physics

(Joyce) #1

6.46. Eb = Re 4 /2h^2 , R = μe 4 /2h 3 , where It is the reduced mass
of the system. If the motion of the nucleus is not taken into account,
these values (in the case of a hydrogen atom) are greater by m/M
0.055%, where m and M are the masses of an electron and a pro-
ton.
6.47. ED — EH = 3.7 meV, — A.D = 33 pm.
6.48. (a) 0.285 pm, 2.53 keV, 0.65 nm; (b) 106 pm, 6.8 eV,
0.243 pm.
6.49. 123, 2.86 and 0.186 pm.
6.50. 0.45 keV.
6.51. For both particles X, = 27th (1 + mn/md)/r2ninT = 8.6 pm.
6.52. I= a 1 A, 2 /Y.24 + A.:.
6.53. X= 27th/ V 2mkT =128 pm.
6.54. First, let us find the distribution of molecules over de
Broglie wavelengths. From the relation f (v) dv = (A.) where
f (v) is Maxwell's distribution of velocities, we obtain


q^ = Ak-4e-a/^12 , a = 2n^2 h^2 / n.kT.
The condition 41(14= 0 provides 4 7 .= Tarr mkT = 0.09 nm.
6.55. X, = 2nh/V2m7' (1 -I- T/2mc 2 ), T < 4mc 2 A?j1 = 20.4 keV (for
an electron) and 37.5 MeV (for a proton).


6.56. T = (j/ 2 1) mc 2 = 0.21 MeV.
6.57. Xonnil mcX.hlah= 3.3 pm.
6.58. v -=4rth//mbAx=-- 2.0-10 6 m/s.
6.59. Ax = 23 - thlIdli2meT/ = 4.9 [tm.
6.60. Vo = n 2 h 2 /2rne (j/TI-1) 2 d 2 sin 2 0 = 0.15 keV.
6.61. c/ = z-ak117 2mT cos (0/2) = 0.21 nm, where k= 4.
6.62. d= Takli 2mT sin 0 = 0.23 ± 0.04 nm, where k = 3 and the
angle 0 is determined by the formula tan 20 =D121.
6.63. (a) n= 111 +1/ 1 /V= 1.05; (b) V/Vi_.>-1/1 (2 -I-1) = 50.
6.64. En = n 22 - c 2 h 2 /2m1 2 , where n = 1, 2,...


6.66. 1.10 (^4) , 1.10 and 1.10-2° cm/s.
6.67. Au






hlml = 1.10 6 m/s; vi = 2.2.10 6 m/s.
6.69. At


im/ 2 /h 10 -16 s.

6.70. Tmin h 2 /2m1 (^2) = 1 eV. Here we assumed that p Ap
and Ax = 1.
6.71. Av/v ti h//1/2mT = 1.10".
6.72. F
6.73. Taking into account that p Ap Ax ,h1x, we get
E = T U h^2 /2mx 2 kx 2 /2. From the condition dEl dx = 0
we find x, and then Emin klm = he), where co is the oscillat-
or's angular frequency. The rigorous calculations furnish the value
ho /2.
6.74. Taking into account that p Ap Ar and Ar r,
we get E = p 2 /2m — e 2 /r h 2 /2mr 2 — e 2 /r. From the condition

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