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Unformatted text preview: ’ Chapter 2 b. 2.;
lgketch tube for photoelectric experiment and in: 14". What V0 is required for 7L = 2440Au
and Pt (4. 09eV) ? (13} Light intensity V0 V Note the same V0 is required
for various intensiﬁes. (c) 2. = 2440 A = 0.244 pm
hv(eV) = 1.24/Mum) == 124/0244 = 5.08 eV
Vo=hv—¢=5.08 4.09~1V (Prob. 2.2
IfVAa = I V, ﬁnd energy and velocity ofe‘ moving from B to A‘ The electron gains 1 eV = 1.6 x 10‘19 J. 2 E: mv (QIH 2 x 1.6 x 10“9
9.1 x 10“31 1/2
v = mm = [ J = 5.93 x 105m/s 13 mwiy«ﬁwwmmwwﬂwﬂ man WWW: 13 Prob. 2.3
Find wavelengths for hydrogen Lyman, Balmer, Paschen series.
(a)
x — E. = __...5__.__.. = 1A2:—
V cR(.L...L) R015" "12)
2 2
"1 "2
2 2 2 2
= 1 ——————”‘ "2 = (911x10‘8cm)——"3—"2——
109,678 (a; — n3) (n3 — n?)
(b) Lymtm
n n2 nzl nzlgnzl) 191111290124;
2 4 3 1.333 1215 a
3 9 8 1.125 1025 :9.
4 16 15 1.067 972 5 25 24 1.042 949 A Limit: 911 21 Similarly, for the Balmex series the limit is 4(9112 = 3644 A, and the wavelengths for n ==
3 to n = 7 are 6559, 4859, 4338, 4100, 3968 A. For the Paschal series the limit is 8199 A and the wavelengths for n = 4 to n =10 are
18741, 12811, 10932, 10044, 9541, 9224, and 9010 A. W
Show Eq. (21 7) corresponds to Eq.(2~3). “mt is, show 4
cR M = 2161121: From the solution of Prob. 2.3, Eq.(217) is 217x10‘191 1 151 1
v =—————————~— =327x10 ____
21 6.63x10'34(n12 21%) (r112 11%)
Eq.(23)is 1,2, = 3x108 x1097x107(——l§~——13)= 3.29x10‘5(12—12~)
n1 n2 n1 "2 14 14 mg ”
'16!) Watimpxifm = 1 A? 110 ~s) 6.63x10‘34((kg  m2/32)s)
. m] = .__.__..... = m
Apxacg s) 21:Ax(m) 21:(10‘l°m) = 1.06x 10‘“ kg mls MathifAE = 1 eV? “ h(eV  s)
At” AE(cV) =6.59x10‘“s ind A for 100eV and 12 Intel/electrons. Comment on e microscopes compared to visible
ight. h h 6.63x10'34
V=425lm, K—= =
, w32E»: [2x E—l/z
mv 9.11x10‘3‘3“2 '.......’ 1/2__  —9 l/2_ 10 _ .
“(Léxlowwzﬂoor ‘123X10 [1001‘ l.23x10 m~1,23A ForleeV: k=1.23xlo‘9[l.2x104]‘"2 = 1.12x10‘“m=0.112A f Visible light is about 0.5m = 5000A, so the resolution of electron microscopy is much
better. The probability of ﬁnding an atom in the unstable state at time t is NMWO = exp(t/'§).
This is analogous to the probability of ﬁnding a particle at x in Eq.(221b). Thus we have Alternative solution: see the approach used in calculating diffusion length, Eqs.(4~37) to
(4.39). 15 15 W
Calculate the expectation value for pi and p, for a plane wave ‘i’ = elk”. jar ” 2
J‘ A‘e~jk.x(§i] Aekaxdx
( 2) 2 km = (Pikx )2 aﬁer normalization because ‘I‘ has no 2 dependence. P b. 2.9
Relate momentum to wave vector fbr a ﬂu: electron described by a plane wave. See Example 31. mm
Calculate the expectation value for px, p}, and E fbr a plane wave Wm) u Adam—7". “f Atemin Max
j 3:: (p3) = “a on
f IAFe’jk‘xejk'xa'x = bk, after normalization 3 [6.63x10’34 (10)”)kgma“ = 1.055x10”33‘kg‘ms‘1 16 16 b. 2.11 Calculate first three energy levels for a 10 A quantum well with inﬁnite walls. From Eq. 2.33 nzxw _ (6.63 x 10"“)2 2 E" : 2mL2 ’ 8 x 9.11 x 1031 >1 {109}2 " B, = 0.603 x 10*19 J = 0.377 eV
E2 = 0.377(4) = 1.508 eV
E3 = 0.377(9) = 3.393 ev Prob. 2.12
Comment on the alkali metals and the halogens. Li, Na and K have one valence 6‘ outside a closed shell. F, Cl and Br require one electron to fill a. shell and to then have electron conﬁgurations
like inert Ne, Ar, Kr. What are the electronic conﬁgurations for IV a+ and Cl“ ? Na+ has 132 252 2p“, which is the [Ne] conﬁguration.
06‘ has 152 237 2p6 332 Spa, which is the [Ar] conﬁguration. 17 17 ...
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This note was uploaded on 03/19/2008 for the course EE 339 taught by Professor Banjeree during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Banjeree

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