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陈宏芳_原子物理学_答案.pdf

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1、1 1.1 13.6eV.(1) (2) 0.6 mm (1) R r 0 V : () R rR V r E dl E dl =+ vv vv 32 00 ( ) 4 4 R rR e re V r dr dr Rr pe pe =+ = 22 3 0 ( 3) 8 eRr R pe - 222 3 0 ( 3) 8 eRr E R pe - = 2 0 3 8 e R pe 13.6eV R 0.16 nm (2) 2 3 0 4 er F R pe = e F mr = ) x ( u m 2 k ) x ( u x m 2 ) x ( u E 2 2 2 2 2 k h h = - =

2、 m 2 k E 2 2 h = 2.12 m V(x) = 1/2mw 2 x 2 (1) (2) 2 ) 2 ( 0 ) ( x m e x u h w - = 2 ) 2 ( 1 2 ) ( x m xe m x u h h w w - = (3) h p xD D w h (1) 22 2 2 du Vu Eu m dx - += h ) x ( Eu ) x ( u x m 2 1 ) x ( u x m 2 2 2 2 2 2 = + - w h (2) 2 ) 2 ( 0 ) ( x m e x u h w - = 0 2 E w = h 1 3 2 E w = h (3) E

3、22 1 2 Emx w = 10 x Dx p Dp 2 2E x mw D= 2 2 2 2 ) ( 2 2 ) ( 2 x m m p m p E D D h = = = 2 E w = h 2.13 2p3/2 1.610 9s, 7 4.11 10 E eV t - D = D h 2.14 1 2 2 2 n n ma 8 h E = n =1 E 1 =38eV n=2 E 2 =151eV n=3 E 3 =340eV 2.15 5000 0.1 q = 90 0 (1 cos ) 0.0024 e h nm mc l l lq D = - =-= 500nm l = 500.

4、0024nm l = 0.01nm l = 0.0124nm l = 2.16 mc c h = l m 2 0 2 4 c m e r e e pe = m e e (1) 0 0 a r a e c l e c h (2) 137 1 4 0 2 = c e h pe a c l r e 53 . 0 4 2 2 0 0 = me a h pe 11 (3) p p 140 MeV/c 2 1 a pe pe l = = = c e me mc a c h h h 0 2 2 0 2 0 4 4 2 2 0 2 0 4 / a pe = = c e a r e h 2 l c= 0.003

5、9 r e= 2.810 6nm = 2.8 fm 3 6 10 4 . 1 - = = c m p p l h nm = 1.4 fm 2.17 l = 1.7510 7 m max / l hc = 7.08eV 7.08eV 2.18 1.810 18W 6000 l = 600010 8 cm, = hc/600010 8 =2.06 eV, W 18 10 8 . 1 - /( eV 10 6 . 1 19 - *2.06)=5.4 2.19 25 0 C 1.67510 27kg 939.5 MeV/c 2 (1) (2) 2.82 NaCl E 298K (3/2)kT= 6.1

6、710 21J=38.5 meV, (1) m 10 458 . 1 mE 2 h 10 - = = l =0.146 nm, (2) q = 15.5 2.20 1 m 0.5 mE 2 h = l R 1 m ) 10 1 ( 4 9 0 2 = pe e E = 1.4410 9eV l =3.23210 5m= 32.3 mm, l (R 0.510 10 m = 0.23nm 3.1 10.2eV, 121.8nm, 12 3.1 10.2eV, 121.8nm, 121.8nm , . 3.2 n = 2 : 2 2 1/2 S s =1/2 l=0 j =1/2 m j= 1/2

7、2 2 1/2 P s =1/2, l=1 j=1/2 m j =1/24 2 3/2 P s =1/2 l=1 j =3/2 m j = 3/2 1/2 3.3 ) 1 cos 3 ( 6 81 1 2 3 / 2 0 2 / 3 0 0 - = - q p a r nlm e a r a u u nlm n l m l : , exp(r/na 0 ) n sinq cosq l e imj m n=3, l=2,m l =0 3.4 : r 4 e U 0 2 pe - = - = 0 2 10 0 * 10 10 dr r R ) r 4 e ( R U pe = dr ) a r 2 exp( r ) 4 e ( a 4 0 0 0 2 3 0 - - pe= 0 0 2 a 4 e pe - 3.5 2s 2p 13 10 - r cm *2 () nl nl P r dr R R r dr = 2s n=2 l =0 P = - - = r 0 2 0 2 0 3 0 r 0 2 20 * 20 dr r ) a r exp( ) a r 2 ( ) a 2 1 ( dr r R R

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