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Unformatted text preview: Chemistry 341 Physical Chemistry I Fall 2011 Problem Set 1 Due: Monday September 5 For this problem set and subsequent problem sets only a certain number of questions need
to be submitted for grading. For this problem set submit your answers to ONLY FIVE
(5) of the following 10 questions. You are still responsible for understanding the ideas
for all of the problems assigned 1. For Cr the work function, 415 , is 4.40 eV. (at) Calculate the kinetic energy of electrons emitted from a chromium surface
when it is irradiated with uV radiation of wavelength 200 nm.
(b) What is the stopping potential, V5, for these electrons? [Ans (a) 2.88 x10” J: 1.80 eV, (b) 1.80V]
aVA': «3i wuri‘: {xi3 ab (0} hug. zlamwl :. _¢
(a; 4: 2a. 3: to‘“ 33,3 Cums no? mg“) (locum 3 (lo'qm u—g NM.
.. (two av) ke = 3 mm. When a clean surface of Ag is irradiated with light of wavelength 230 nm, the stopping potential, Vs, of the ejected electrons is found to be 0.80 V. Calculate
for Ag: (a) the work function, ¢, and
(b) the threshold ﬁequency, v0. [Ans. (a) 4.59 eV, (b) v0=1.11x1015 Hz] (if: in; _ eV“
7t
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“‘3 In the derivation of the Planck black body formula, Planck’s principal assumption
was that the energies of the oscillators making up the walls of the cavity can have
onlythevalues 8: nhV andthat A8: hv. As v—>0 then A£—>0 and 8 is essentially continuous. Thus we should expect the nonclassical Planck
distribution to go over to the classical Rayleigh—Jeans distribution _ 8rrkT pV(V3D_ C3 V at low frequencies, where A 8 —>0. Show that the Planck black body distribution
function [eqn (9.183) RIO/J) : 8ﬂhh—EV/ C)
W} i reduces to the Rayleigh—Jeans distribution function as v—>0.
2 [NOTE ex =1+x+x—+...] 2 2!
 .. '5
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nee 390 { Qb‘t 1‘} C.5
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5
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1/ 1
= ‘31}; g u o:\ 1m :k
c5 g
W k7 a)?“ g
"5% Win We A"54"“E"J\—A"."‘5"A alA—éwtec’. :5 ’l‘ae BKLAAQLSM— jeans
a.L$+r'.\ou.*m nine,“ ckthanJ m lai bu,“ W “much”
Making. "up +30. ﬁche. enclosing) ‘H’ig awn1r cure .{lnll’ouJ‘uLg
engaged Wtchqnth... Clint/Na, Before Planck’s theoretical work on blackbody radiation, Wien showed
empirically that AMT : 2.897 x 10'3 mK
Where Am is the wavelength at which the blackbody spectrum has its maximum value at a particular temperature T. The expression is called the Wien
displacement law. Derive it from the Planck distribution function expressed in
terms of xi [eqn 9.185] 8 7: h c pi(/1,D=—hc%
2’5 [elkT_ [HJNTz Set X = hc / [tum [CT and derive the intermediate result e”" + (x / 5) = 1. This equation cannot be solved analytically but must be solved numerically.
Solve this by iteration and show that x=4.965 is the solution] gkc‘kaum : wry—ﬂ ass€31.14 3k : AX
{exll “m whom w'\\:\ 0.3ch where bark—(L A5 92—N‘eac g °\§,‘ ._
m 0 So ‘3
A .3! qu .. X ex 3 1 c:
(e‘ ~11 @xquz AX“ {50;va — xex} =0
(exHZ 'Tke china memwvuﬁ—Qvat Sc\u,\—m {3 X
'5 (aw—H xe. :o Mwl+lp‘ﬂ? Luz, ef" dual c)‘\v\0\e, lows 4% 39:" x
L—e Z..=o
b Cow‘ﬁnueoK 3 Us‘woa EXCEL , we {Sn/LA, by) “LVLJ am} arrow +exm+ 44:3
so Lug01A A": 'ﬁﬁis L\u_a_5anA (“lb \{sieam ¥.c,a..u.+ {LLquY‘eS\ 'LS mm = mxmmagsm
3‘me = 1019...—
xv,“ kr w _
Cbeszm 3 A) (2.“.ng lucbmgq
Cwqu {1.39m “34% :10“) T %
NWT = km“ —. The theory of blackbody radiation is used regularly in astronomy to estimate the
surface temperature of staxs. Use the Wien formula of question 4 to estimate the
surface temperature of (a) the sun Where Am is 500 nm and (b) the star Sirius, which is one of the hottest known stars and appears blue,
Where 2mm is 260 nm. 3m; '1' . 2.2% x if" mK (mum's Lew"? ‘T ': 2.9,om 1. LquM \< \Max («A "Fete sun 7mm“; :gobnm T: 2BQ1xxo.3vvuK , ( nm )
(500mm) luto'qm
3 T= ‘C W Cm +9.2 «,A—M $\2\U$ “T”: 39393? x 1&3 _ mm 3
(manna/q \XIO‘Q‘M mmww‘mm‘ “aw 6. SAB 9.5, p.344.
[Ans (a) 1879 x 10—2 nm, (b) 1.240 nm] 9.5 Electrons are accelerated by a 1000V potential drop. (0)
Calculate the do Broglie wavelength. (b) Calculate the wave—
length of the Xrays that could be produced When these elec—
trons strike a solid. (63 «EN = «%x ‘q "2.
: z (able.qu k3) (Lcozxm C3 (mom/3 k M‘s}
g. 4T 3
P: x 10?, x. 16—23 k5 MA.“ X: (then: x 15$“ 3A3 _ (kcpml‘iz)
(“7082; Led} 3 ‘\ l: 3.%‘)Q3 yo’ M ) __ é .Q'jqﬁ 5:, Leah“ M\( m 3 : ??‘%7q ’5 \ x ‘0  q M we ﬁE ‘5‘4 
“R : (LL20, xv: 3143 (z‘iﬁ%x Lo‘bML‘X ( “W‘ 61.902 ,‘ ml‘icﬁ CmooV) (i
I :v ix “3‘9 w. k = “2‘4 . WW ,. N w. SAB 9.9, p. 344.
[Ans (a) 2.21 x 10'33 m, (b) 6.63 x 10‘19 m, (c) 6.63 x 10—11m,
(d) 3.979 x 10'10 m] 9.9 Calculate the de Broglie wavelengths of the following:
(a) A 1—g bullet with velocity 300 In 5‘1
(b) A 10‘6g particle with velocity 10—6 m s'1
(c) A 10—10g particle with velocity 10‘10 m 51
(d) An H2 molecule with energy of ng at T = 20 K C03 k“: g;
p
3‘: L
ww
)\ : (6.6.7.03. x63“ 3AA . (13 M2; )
Choc x u)“; kg) (3 com") IT (bx >\ = CQ.GZB 1 Lo‘g‘i—TAQ ,< kg Mtg}
(_.oox \o‘q kg) (nooxwﬂbmit (:3 >\: (Ca(91b x \DJ’RUQ ‘ (k3 “21:13
(i.0§x m‘ﬂ’kc‘rﬁ (Loo >6 iu'mmﬂ‘3 j 1: (wrogx‘bo‘nm z 7. . v: {}:k::}llz : g.3KI\‘EE.}tl'2. '/
9: @6653} 7'
(2.0: S 91% x to“ k3wurQ'l) U‘: “NFL; Mg" 3‘: (Lulfox \53H3A\ his W3. 4.)
(1.039 8%x 1J5 hydra9") (“RISmi‘B' 3' Derive the expression for the energy of a particle in a 1—D box by using the de
Broglie formula and the condition for a standing wave in 1D box. 7x . L C0 32— Efagl‘; reiLJtVpiau‘ i u:er nrI‘L... C23 calihjm 'r$+n_ A.‘ H
A? r I) .y > "D WLVE. A 1% 4% t... L732 L:)§ 7— _— 7. For a particle in a 1D box the ground state (n=l) wavefunction is Hamel?) Determine: (a) the probability of ﬁnding the particle between a/4 —> 3a/4 and (b) the probability of ﬁnding the particle between O—> a/3. Comment on your ﬁndings as to What you would expect classically in parts (a) and (b).
[Ans. (a) 0.8183, (b) 0.1955] 343;.lul
q
¥=°~lq
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= T
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9. 3 
V(q—>ﬁ°~ owe?) using $UENTW—1c megFLA"? “’3 ‘0‘.“ r l 37r/4
3 (2M)! sinzydy= %(%+%n) = .81831 7r/4
“35
(b3 2 2
Ho» 2:9 Ac <‘e a4
ads 1 M
" ' 3‘... “M T
Q Ezb '31,, P(°>°:\ ‘ gtﬁ ‘3’ Mwm‘zﬂwc u = _§b m©3%[o\}1 u : “L. _ FL ‘
s Tr m a = a; ~ :0.me
pko‘aﬁ : ‘ ‘ [—31— ‘ Sawﬂy».ng
1; s x ULsmcal ‘bchrHW—m w; c»ng kME J No; Line}. § 5 (2/n)j:’3sin2ydy = %(—%ﬁ + g—n) = .1955 (63 Far 0. £10551 co..\ Pa—ere , feta9 72%) = 3?: whereas have (‘c‘f He OEY‘uULhc‘ “n+4. A'éﬁqv 7 a (L For 1 c\o.sS\m.\ Vean—iclg) ? (3"?S‘gxg—‘i whzre 6.8 ‘09“: Lu: 4{a °3~mumA sable @(bwaeej’) 4. l
3 an . 3 10. («A 0% Consider a particle in a lD box of length a in state n=5 . (a) What are the most likely locations of the particle? (b) When a transition from n=5—>n=6 occurs an energy separation AE results.
Calculate the energy separation in units of J , eV, cm‘l, and kJ 11101’1 for an
electron in a 1Dbox With length a=1.50 nm. [Ans. (a) a/lO, 3a/10, a/2, 7a/10, 9a/10, (b) 2.945 x 10‘19 J, 1.839 ev, 1.483 x104
cm‘l, 177.4 k] moi—1] "322.
1/5 ‘43! 315 415
o a" 1
m x . ~ '1 q
“‘M‘LF“ a" 15: a 131. >51; 311%;  M: = E 3110.? m‘} = a 4L2
8MQ7‘ 3M4}—
AE‘: N (to ~626x Le'qu/LDE' I _ he MZA'I
3 CQJOQx 15%}:3) (\S'ox\o‘q wt} 3’ ‘ AEl ‘1 HA' I: (BIO \ 1'5 L‘)< _C‘\.A5>l.\0L 05S\
.e 2 I
At‘ .— 2 ‘  ...
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This note was uploaded on 01/23/2012 for the course CHM 341 taught by Professor Klier during the Fall '08 term at Lehigh University .
 Fall '08
 KLIER

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