Physics exercises_solution: Chapter 41

Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 41 | 2 T h i 7 7 1A L 2 x 10-34 J s I L Jl l 1 n l l 1 1-1 I in_34 T I qJ 10 34 J sJ l l 1 l 4. a m 2 so L 2q. b y l l 1 r 76-q . c The angle is arccc arccos m and the angles are for mt -2 to mt 2 L J U6J . The angle corresponding to mt l will always be larger for larger l. L jl l 1 r . The maximum orbital quantum number l n -1. So if n 2 l 1 L - l 1 n 72 n 20 l 19 L 719 20 n n 200 l 199 L 7199 200 n The maximum angular momentum value gets closer to the Bohr model value the larger the value of n. The l mt combinations are 0 0 1 0 1 1 2 0 2 1 2 2 3 0 3 1 3 2 3 3 4 0 4 1 4 2 4 3 and 4 4 a total of 25. X . t . x eV b Each state has the same energy n is the same ---------25 eV. 415 u 0 -1 x 10-19C 2 x t-l-. J . .3 X 4ne0 r 4ns0 x 10-10 m x 10-18J x 10-19 J eV eV. a As in Example the probability is 2 pa 2 - - 4 P 1 I V1S I 4nr dr j0 a ar a2r 3 2rja a 2 2 2 4 e 0 . 5e 1------- . 2 b The difference in the probabilities is 1 - 5e J - 1 - V2 e-1 2 e - 2e J . a y 2 y y R r 2 0 2 Ae- m Ae m A2 R r 2 3 2 which is independent oft b e 2n n n f2n n 1 0 0 2 def A2 def 2nA2 1 A -j 1 me4 E. 41 E __ __r__ e E - E 1- E - 0 75 E n 4n 0 2 2n2n2 12 2 1 22 1 p a If mr m x 10 3 kg mre 4 9. 09 X 10-31kg 2 X 10 Cl x 109 Nm2 c 2- 4ne0 2n2 2 x 10 J s 2 V 7 x 10-18 J eV. For 2 1 transition the coefficient is eV eV. m b If mr using the result from part a m e eV fmi f3 59 eV eV. 4ne0 2n I m 2 . . . f eV _ r Similarly the 2 1 transition I-2-I eV. c Ifmr using the result from part a eV ft858 2525 eV 4ne0 2 n2 m and the 2 1 transition gives eV 1893 eV. A 0hh 8 X10-34J s 2 a m m a. 0 -------------0---------------------- - 1 nme2 n x 10-31 kg x 10-19C 2 a x 10-11m b mr m a2 2a x 10-10 m. 1 13 c m m a .