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phys 2D

# phys 2D - 6-6 x = A cos kx B sin kx = \$ kA sin kx kB cos...

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6-6 " x ( ) = A cos kx + B sin kx # # x = \$ kA sin kx + kB cos kx # 2 # x 2 = \$ k 2 A cos kx \$ k 2 B sin kx \$ 2 m h 2 % & ( ) * E \$ U ( ) = \$ 2 mE h 2 % & ( ) * A cos kx + B sin kx ( ) The Schrödinger equation is satisfied if " 2 # " x 2 = \$ 2 m h 2 % & ( ) * E \$ U ( ) or " k 2 A cos kx + B sin kx ( ) = " 2 mE h 2 # \$ % & ( A cos kx + B sin kx ( ) . Therefore E = h 2 k 2 2 m . 6-9 E n = n 2 h 2 8 mL 2 , so " E = E 2 # E 1 = 3 h 2 8 mL 2 " E = 3 ( ) 1240 eV nm c ( ) 2 8 938.28 # 10 6 eV c 2 ( ) 10 \$ 5 nm ( ) 2 = 6.14 MeV = hc # E = 1240 eV nm 6.14 \$ 10 6 eV = 2.02 \$ 10 % 4 nm This is the gamma ray region of the electromagnetic spectrum. 6-10 E n = n 2 h 2 8 mL 2 h 2 8 mL 2 = 6.63 " 10 # 34 Js ( ) 2 8 9.11 " 10 # 31 kg ( ) 10 # 10 m ( ) 2 = 6.03 " 10 # 18 J = 37.7 eV (a) E 1 = 37.7 eV E 2 = 37.7 " 2 2 = 151 eV E 3 = 37.7 " 3 2 = 339 eV E 4 = 37.7 " 4 2 = 603 eV (b) hf = hc = E n i # E n f = hc E n i # E n f = 1240 eV \$ nm E n i # E n f For n i = 4 , n f = 1 , E n i " E n f = 603 eV " 37.7 eV = 565 eV , = 2.19 nm n i = 4 , n f = 2 , = 2.75 nm n i = 4 , n f = 3 , = 4.70 nm n i = 3 , n f = 1 , = 4.12 nm n i = 3 , n f = 2 , = 6.59 nm n i = 2 , n f = 1 , = 10.9 nm

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6-12 " E = hc # = h 2 8 mL 2 \$ % & ( ) 2 2 * 1 2 [ ] and L = 3 8 ( ) h " mc # \$ % % & ( ( 1 2 = 7.93 ) 10 * 10 m = 7.93 Å. 6-13 (a) Proton in a box of width L = 0.200 nm = 2 " 10 # 10 m E 1 = h 2 8 m p L 2 = 6.626 " 10 # 34 J \$ s ( ) 2 8 1.67 " 10 # 27 kg ( ) 2 " 10 # 10 m ( ) 2 = 8.22 " 10 # 22 J = 8.22 " 10 # 22 J 1.60 " 10 # 19 J eV = 5.13 " 10 # 3 eV (b) Electron in the same box: E 1 = h 2 8 m e L 2 = 6.626 " 10 # 34 J \$ s ( ) 2 8 9.11 " 10 # 31 kg ( ) 2 " 10 # 10 m ( ) 2 = 1.506 " 10 # 18 J = 9.40 eV . (c) The electron has a much higher energy because it is much less massive. 6-14 (a) Still, n 2 = L so p = h = nh 2 L K = c 2 p 2 + mc 2 ( ) 2 " # \$ % & ( mc 2 ( ) = E ( mc 2 E n = nhc 2 L ) * + , - . 2 + mc 2 ( ) 2 " # \$ \$ % & , K n = nhc 2 L ) * + , - . 2 + mc 2 ( ) 2 " # \$ \$ % & ( mc 2 (b) Taking L = 10 " 12 m , m = 9.11 " 10 # 31 kg , and n = 1 we find K 1 = 4.69 " 10 # 14 J . The nonrelativistic result is E 1 = h 2 8 mL 2 = 6.63 " 10 # 34 J \$ s ( ) 2 8 9.11 " 10 # 31 kg ( ) 10 # 24 m 2 ( ) = 6.03 " 10 # 14 J Comparing this with K 1 , we see that this value is too big by 29%.
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phys 2D - 6-6 x = A cos kx B sin kx = \$ kA sin kx kB cos...

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