# Homework#7 - —2\II\I —— AQr2 exp C 0° 0° —2 2 A2043 ‘l\Ildr=A2 r2exp dr=A2 3 =:1 0 0 a(2/00 4 Therefore A = —4—3 = 2044/2 CY 8 I E

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Unformatted text preview: / —2 \/ \II*\I/ —— AQr2 exp C! 0° 0° —2 2 A2043 ‘l/*\Ildr=A2/ r2exp dr=A2[ 3] = :1 0 0 a (2/00 4 Therefore A = —4—3 = 2044/2 CY 8. I E _ 3.4 + 3.9+5.2 + 4.7+4.1 + 3.8+ 3.9 +4.7+4.1 + 4.5 + 3.8+4.5 +4.8+3.9+4.4 ' 15 = 4.247 <w2> = - 3.42 + 3.92 + 5.22 + 4.72 + 4.12 + 3.82 + 3.92 + 4.72 + 4.12 + 4.52 + 3.82 + 4.52 + 4.82 + 3.92 + 4.42 15 (\$2) = 18.254 The standard deviation is a: min—a2: 2(x3-2wm‘f2): N N Look at the three terms in the sum. The ﬁrst is just The second is —2?I:‘ 2 ~25? The third term is Zr? 2 N552 2 a? N N Putting the results together a = (:32) —— 23132 +52 = (332) — E? For the data given we have a = was?) — a? = M18254 — (4.247)2 = 0.466 . * 10. l;Using the Euler relations between exponential and trig functions 1? = A (6“: + 6'”) = 2A cos Normalization: Tr 7r WW dm 2 4142/ cos2 d3: = 4A27r = 1 —'7|' —7T 1 . . . . . . Thus A = Zﬁ and the probability of bemg 1n the interval [0,7r/8] 1s 7T/8 l 7r/8 1 Tr/B l P = 1/)*1/Jda: = —/ cos2 dx = - E + ~1—sin(2rc) 0 7r 0 7r 2 4 O 1 1 = — + = 0.119 12. 2w2h2 (n +1)2 7r2,32 I E” 2mL2 En“ _ 2mL2 7T2h2 2 W252 AEn = n — E. = — 2 = E +1 2ng [(n+ 1) n my (2n+ 1) Computing speciﬁc values 71.2712 AE1 -— 2mL2 (3) L 7r2h2 } «2712 AEgoo = mLQ (1601) / * 15.: a) Starting with Equation (6.35) and using the electron mass and the length given, we have 2722 7r2(hc)2 E = 2 7T : 2 n n 2mL2 n 2(7TLC2)L2 2 9 .3 V- 2 n2 w 2 H2 (940 X IO—sev) 2(5.11 >< 105eV) (2000 nm) 3 7‘; «)1 Then the three lowest energy levels are: E1 = 9.40 x 10‘8 eV; E2 =lirrj8‘8 >< 10~7 eV; and E% 32782 x 10—7ev; 351\$ 3 3 16V b) Average kinetic energy equals EkT = 5 (1.381 x 10“23 J/K) 13K which equals 1.68 x 10‘3 eV. Substitute this value into the equation above as E,1 and solve for n. We ﬁnd n = 134. 22. Lacking an explicit equation for finite square well energies7 we will approximate using the inﬁnite square well formula. In order to contain three energy levels the depth of the well must be at least Eﬁ ugh? _. 9h2 — 8an2 ’ 8mL2 Evaluating numerically with the given mass 2 2 2 I 2 9h : 9h 0 9(1240 eV nm) I ~96'1MQV : 8mL2 8mch2 — 8(2 X 109 eV) (3 X 10#6 nm)2 E 233,211) The wavelengths are longer for the ﬁnite well, because the wave functions can leak outside the box. b) Generally shorter wavelengths correspond to higher energies, so we expect energies to the lower for the ﬁnite well. c) Generally the number of bound states is limited by the depth of the well. We expect no bound states for E > V0. 30. 1/10(:B) has these features: it is symmetric about at = 0; it has a maximum at :r = 0 because "(the Wave function must tend toward zero for :L' —+ ioo; there is no node in the ground state; the wave function decreases exponentially where V > E. * 31. AEn=En+1—En=<n+1+é>hw—(n+%)hw=hw foralln This is true for all n, and there is no restriction on the number of levels. 1 ' E = (n +7111) = (71+ 5) hf = (4.136 ><10_15 eV . s) (1013 s'l) <n+ = (4.136 x 10*2 eV) (71+ For the harmonic oscillator 0J2 = —— so m k = 102m = 47?me = 4m? (1013 3—1)2 (3.32 x 10‘26 kg) = 131 N/m ...
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## This note was uploaded on 04/23/2011 for the course PHYS 342 taught by Professor Staff during the Spring '08 term at Purdue University-West Lafayette.

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Homework#7 - —2\II\I —— AQr2 exp C 0° 0° —2 2 A2043 ‘l\Ildr=A2 r2exp dr=A2 3 =:1 0 0 a(2/00 4 Therefore A = —4—3 = 2044/2 CY 8 I E

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