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ssm_ch39 - Chapter 39 13 R The probability that the...

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Chapter 39 13 The probability that the electron is found in any interval is given by P = R j à j 2 dx , where the integral is over the interval. If the interval width ¢ x is small, the probability can be approximated by P = j à j 2 ¢ x , where the wave function is evaluated for the center of the interval, say. For an electron trapped in an infinite well of width L , the ground state probability density is j à j 2 = 2 L sin 2 ³ ¼x L ´ ; so P = μ 2 ¢ x L sin 2 ³ ¼x L ´ : (a) Take L = 100pm, x = 25 pm, and ¢ x = 5 : 0pm. Then P = · 2(5 : 0 pm) 100 pm ¸ sin 2 · ¼ (25 pm) 100 pm ¸ = 0 : 050 : (b) Take L = 100pm, x = 50pm, and ¢ x = 5 : 0pm. Then P = · 2(5 : 0 pm) 100 pm ¸ sin 2 · ¼ (50 pm) 100 pm ¸ = 0 : 10 : (c) Take L = 100pm, x = 90 pm, and ¢ x = 5 : 0pm. Then P = · 2(5 : 0 pm) 100 pm ¸ sin 2 · ¼ (90 pm) 100pm ¸ = 0 : 0095 : 25 The energy levels are given by E n x n y = h 2 8 m " n 2 x L 2 x + n 2 y L 2 y # = h 2 8 mL 2 " n 2 x + n 2 y 4 # ; where the substitutions L x = L and L y = 2 L were made. In units of h 2 = 8 mL 2 , the energy levels are given by n 2 x + n 2 y = 4. The lowest five levels are E 1 ; 1 = 1 : 25, E 1 ; 2 = 2 : 00, E 1 ; 3 = 3 : 25, E 2 ; 1 = 4 : 25, and E 2 ; 2 = E 1 ; 4 = 5 : 00. A little thought should convince you that there are no other possible values for the energy less than 5. The frequency of the light emitted or absorbed when the electron goes from an initial state i to a final state f is f = ( E f ¡ E i ) =h and in units of h= 8 mL 2 is simply the difference in the values of n 2 x + n 2 y = 4 for the two states. The possible frequencies are 0 : 75 (1,2 ¡! 1,1), 2 : 00 (1,3 ¡! 1,1), 3 : 00 (2,1 ¡! 1,1), 3 : 75 (2,2 ¡! 1,1), 1 : 25 (1,3 ¡! 1,2), 2 : 25 (2,1 ¡! 1,2), 3 : 00 (2,2 ¡! 1,2), 1 : 00 (2,1 ¡! 1,3), 1 : 75 (2,2 ¡! 1,3), 0 :
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