CH8 - 8-1 2 2 2 2 2 n1 n2 n 3 E 2 m Lx Ly L z 22 2 2 2 Lx L...

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Unformatted text preview: 8-1 2 2 2 2 2 n1 n2 n 3 E 2 m Lx Ly L z 22 2 2 2 Lx L , Ly Lz 2L . Let 2 E0 . Then E E0 4n 1 n 2 n 3 . Choose the quantum 8mL numbers as follows: n1 n2 n3 E E0 1 1 1 2 1 2 2 2 1 1 1 2 1 1 2 1 2 2 1 3 1 1 2 1 2 2 1 2 3 1 6 9 9 18 12 21 21 24 14 14 * * * ground state first two excited states next excited state * * next two excited states Therefore the first 6 states are 111 , 121 , 112 , 122 , 113 , and 131 with relative energies E 6 , 9, 9, 12, 14, 14. First and third excited states are doubly degenerate. E0 8-2 (a) n1 1 , n2 1 , n3 1 3 6.626 10 34 Js 3 2 2 3h2 E0 2mL2 8 mL2 8 9.11 10 31 kg 2 10 10 m 2 2 4.52 10 18 J 28.2 eV (b) n1 2 , n2 n1 1 , n2 n1 1 , n2 2 6h E1 2 8mL 1 , n 3 1 or 2 , n 3 1 or 1 , n3 2 2E0 56.4 eV 8-3 n 2 11 (a) (b) 2 2 2 11 2 2 E n 2 2 2 mL2 mL n1 1 1 3 n2 1 3 1 n3 3 1 3-fold degenerate 1 (c) x y 3 z 113 A sin sin sin L L L x 3 y z 131 A sin sin sin L L L 311 A sin 8-4 (a) 3 x y z sin sin L L L x , y 1 x 2 y . In the two-dimensional case, Asin k1 xsin k 2 y where n1 n 2 k1 L 2 2 and k2 2 L . (b) E n1 n 2 2 mL 2 2 22 If we let E0 2 , then the energy levels are: mL n1 1 1 2 2 8-5 (a) n2 1 2 1 2 E E0 1 5 2 5 2 4 11 12 21 22 doubly degenerate n1 n 2 n 3 1 and E111 3 6.63 10 34 3 h2 2.47 10 13 J 1.54 MeV 8mL2 8 1.67 10 27 4 10 28 2 (b) 22 12 12 h2 2E States 211, 121, 112 have the same energy and E and states 221, 122, 212 have the energy E 8mL 22 22 12 h2 2 111 3.08 MeV 8mL 2 3E111 4.63 MeV . (c) 8-6 Both states are threefold degenerate. be zero. Substituting r, t 1 x 2 y 3 z t into Schrödinger’s equation with U r 0 2 2 2 2 gives r , t i r , t. Upon dividing through by 2m x 2 y 2 z 2 t 2 1x 2y 3z i t 1 x 2 y 3 z t we obtain . Each term in this 2m 1 x 2 y 3 z t equation is a function of one variable only. Since the variables x, y, z, t are all independent, each term, by itself, must be constant, an observation leads to the four separate equations There is no force on a free particle, so that U r is a constant which, for simplicity, we take to 2 1x E1 2m 1 x x 2 x E2 2m 2 2 2 x 3 x E3 2m 3 t i E t This is subject to the condition that E1 E2 E3 E . The equation for 1 can be rearranged d2 1 2mE1 x , whereupon it is evident the solutions are sinusoidal as 2 2 1 dx 2mE 1 x 1 sin k1 x 1 cos k1 x with k12 2 1 . However, the mixing coefficients 1 and 1 are indeterminate from this analysis. Similarly, we find 2 y 2 sin k 2 y 2 cos k 2 y 3 z 3 sin k3 z 3 cos k 3 z 2mE 2 2mE 3 2 2 and k3 with k2 2 2 . The equation for can be integrated once to get E t e i t with and another indeterminate coefficient. Since the energy operator is E i and i E energy is sharp at the value E in this state. Also, since t t 2 2 2 2 px 2 2 and 2 2 1 k1 1 the magnitude of momentum in the x direction x x is sharp at the value k1 . Similarly, the magnitude of momentum in the y and z directions are sharp at the values k2 and k3 , respectively. (The sign of momentum also will be sharp here if the mixing coefficients are chosen in the ratios 8-10 n 4 , l 3 , and m l 3 . 1 i , and so on). 1 (a) (b) L ll 1 33 1 2 3 3.65 10 12 12 34 Js 34 Js Lz ml 3 3.16 10 8-12 1 1 r e r a 0 a0 12 32 (a) ( r) r (b) The probability of finding the electron in a volume element dV is given by dV . Since the wave function has spherical symmetry, the volume element dV is identified here with the volume of a spherical shell of radius r, dV 4 r 2 dr . The probability of finding the electron between r and r dr (that is, within the spherical shell) is P dV 4 r dr . 2 2 2 2 (c) P r = a0 2 2 r 1 1 a0 0 4 a0 0 (d) 2 2 r a 2 2 r a 2 dV 4 r dr 4 3 e 0 r dr 3 e 0 r dr Integrating by parts, or using a table of integrals, gives 3 4 a0 3 2 1 . 2 dV 3 2 a a0 0 2 r2 r1 (e) 2 P 4 r dr where r1 2 a0 3a and r2 0 2 2 4 r2 P 3 r2 e 2 r a 0 dr a0 r1 let z 2r a0 1 3 2 z z e dz 2 1 3 1 z 2 2 z 2 e z integrating by parts 1 2 17 5 e 3 e 1 0.496 2 2 8-13 Z 2 for He (a) For n 3 , l can have the values of 0, 1, 2 l 0 ml 0 l 1 m l 1, 0, 1 l 2 m l 2, 1, 0, 1, 2 (b) All states have energy E3 Z 2 3 2 13.6 eV E3 6.04 eV . 8-14 Z 3 for Li 2 (a) (b) n 1 l 0 ml 0 n 2 l 0 ml 0 and l 1 m l 1, 0, 1 3 2 For n 1 , E1 2 13.6 122.4 eV 1 32 For n 2 , E2 2 13.6 30.6 eV 2 8-16 For a d state, l 2 . Thus, m l can take on values –2, –1, 0, 1, 2. Since Lz ml , Lz can be 2 , , and zero. (a) For a d state, l 2 8-17 L ll 1 6 12 12 1.055 10 34 Js 2.58 10 34 Js 1.055 10 34 Js 3.65 10 34 Js (b) For an f state, l 3 L ll 1 12 12 12 8-18 The state is 6g (a) n6 (b) (c) En 13.6 eV 2 n E6 13.6 eV 0.378 eV 2 6 For a g-state, l 4 L ll 1 4 5 20 4.47 12 12 (d) m l can be –4, –3, –2, –1, 0, 1, 2, 3, or 4 Lz ml ml 12 Lz ml ; cos L 20 ll 1 ml 4 3 2 1 0 1 2 3 4 Lz 4 3 2 0 2 3 4 153.4 132.1 116.6 102.9 90 77.1 63.4 47.9 26.6 8-21 (a) 2 s r 1 42 12 1 r r 2 a 0 e . At r a0 0.529 10 10 m we find 2 a0 a0 1 4 2 12 32 2 s a0 1 a 0 32 2 1e 1 2 1 0.380 a0 9.88 10 14 32 1 0.380 10 0.529 10 m 32 m 3 2 (b) (c) 8-22 2 s a0 9.88 1014 m 3 2 2 2 9.75 10 29 m 3 2 2 10 Using the result to part (b), we get P sa0 4 a0 2 sa0 3.43 10 2 1 26 12 m 1 . R2 p r Are r 2a 0 where A a0 52 2 Pr r2 R2 p r A 2 r 4 e r a 0 r rP rdr A 2 r 5e r a 0 dr A 2 a6 5! 5a0 2.645 Å 0 0 0 ...
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This note was uploaded on 04/12/2011 for the course PHYS 2D taught by Professor Hirsch during the Winter '08 term at UCSD.

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