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\/ \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
($2) = 18.254 The standard deviation is a: min—a2: 2(x32wm‘f2):
N N Look at the three terms in the sum. The ﬁrst is just The second is —2?I:‘ 2 ~25?
The third term is Zr? 2 N552 2 a?
N N
Putting the results together
a = (:32) —— 23132 +52 = (332) — E? For the data given we have a = was?) — a? = M18254 — (4.247)2 = 0.466 . * 10. l;Using the Euler relations between exponential and trig functions 1? = A (6“: + 6'”) = 2A cos Normalization: Tr
7r
WW dm 2 4142/ cos2 d3: = 4A27r = 1
—'7' —7T
1 . . . . . .
Thus A = Zﬁ and the probability of bemg 1n the interval [0,7r/8] 1s
7T/8 l 7r/8 1 Tr/B l
P = 1/)*1/Jda: = —/ cos2 dx =  E + ~1—sin(2rc)
0 7r 0 7r 2 4 O
1 1
= — + = 0.119 12.
2w2h2 (n +1)2 7r2,32 I
E” 2mL2 En“ _ 2mL2
7T2h2 2 W252
AEn = n — E. = — 2 =
E +1 2ng [(n+ 1) n my (2n+ 1)
Computing speciﬁc values
71.2712
AE1 — 2mL2 (3) L
7r2h2 }
«2712
AEgoo = mLQ (1601) / * 15.: a) Starting with Equation (6.35) and using the electron mass and the length given, we have 2722 7r2(hc)2
E = 2 7T : 2
n n 2mL2 n 2(7TLC2)L2 2 9 .3 V 2
n2 w 2 H2 (940 X IO—sev)
2(5.11 >< 105eV) (2000 nm) 3 7‘;
«)1 Then the three lowest energy levels are: E1 = 9.40 x 10‘8 eV; E2 =lirrj8‘8 >< 10~7 eV; and
E% 32782 x 10—7ev; 351$ 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 UniversityWest Lafayette.
 Spring '08
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