chapter-2-solutions - ’ Chapter 2 b. 2.; lgketch tube for...

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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 nz-l nzlgnz-l) 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. (2-1 7) corresponds to Eq.(2~3). “mt is, show 4 cR M = 2161121: From the solution of Prob. 2.3, Eq.(2-17) is 217x10‘191 1 151 1 v =—————————~—- =327x10 ____ 21 6.63x10'34(n12 21%) (r112 11%) Eq.(2-3)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 ‘1-23X10 [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.(2-21b). 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 3-1. mm Calculate the expectation value for px, p}, and E fbr a plane wave Wm) u Adam—7". “f Ate-min Max j 3:: (p3) = “a on f IAFe’jk‘xejk'xa'x = bk, after normalization 3 [6.63x10’34 (10)”)kg-ma“ = 1.055x10”33‘kg‘m-s‘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 10-31 >1 {10-9}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.

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chapter-2-solutions - ’ Chapter 2 b. 2.; lgketch tube for...

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