Phys chem II CHM4411 Exam 1 Solutions

Phys chem II CHM4411 Exam 1 Solutions - CHM4411—02...

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Unformatted text preview: CHM4411—02 [Phys-Chem. II], Spring 2012 Instructor: H. Mattoussi (6 pages total) Exam 1 (Monday Jan. 30, 2012) _____ “Ki- You name; Exercise 1 (6 points) Prior to Planck’s derivation of the energy distribution law for black—body radiation, Wien found empirically a closely related distribution function that is very nearly but not exactly in agreement with the experimental results, namely: pOtJT) = (ea/ks) >< exp(—b/7thT). This formula shows small deviations from Planck’s expression at long wavelengths. 8nhc hc Afem ~l] (a) By fitting (comparing) Wien’s empirical formula to Planck’s at short wavelengths determine the constants a and Z7. ' (b) Demonstrate that Wien’s formula is consistent with Wien’s law (XmaxXT = constant). Planck’s distribution function is given by: ppmnck (AT) : Answers: (a) We first need to simplify the Planck expression at small X. For 7t small (hc/kkBT) >>l. This he hc also implies that we can use the transformation: (emT — l) = e’V‘BT hc 8nhc MET A~5 This expression is identical to that of Wien with a = 87thc and b = hc The Planck formula can then be written as: ppmck (LT) = e (b) Wiens’ formula is consistent with Wiens’ law, because: At kmax the distribution function must satisfy the differential equation: [98] :0 6/1 H max We; maxi 4:9 I712, f6?” Aidan “AV/waif A A Zesl Q d {shah Q Q/_ r N 3-3 a A >3 *‘L W “Jaw-MN; X a ( {45% 4; 652 (lava) Exercise 2 (4 paints) The work function for a metal Mediated by an incident light signal is 2.09 eV. Calculate the kinetic energy and speed of the electrons ejected by an incident light of wavelength equal to: (a) 1000 nm (b) 300 nm 7 , l f , /.i ..-' i ,- :4" a r; gm ,5? 3/; 4; 15/ ’54; 3 {L M 3*}; an” {:23} iii 5%) LPV‘Qf/“J ‘ k ‘ i Exercise 3 (5 points) An un—normalized Wavefunction ‘P for an electron in a carbon nanotube of length L is: (IJ(X) : sin(2xn/L). Normalize this function. This implies finding the corresponding normalized function ‘PN(X). Hint: sin2(y) =(1~cos(2y))/2 7f? 5/ ,3 L We .94 ea, :1 ’"f‘ fig. 71%” 4‘91) ; oer/“crew fl» V“? 1 x 3 L g" 2“ 27; x, 2 [K73] it. C» K k? ,. ig’i “Tag J i»; X, (ix/,9 p « Exercise 4 (5 points) To provide a physical sense to the idea proposed by de Broglie, that to a moving particle we can also associate a wave with a wavelength A = h/p = h/mv (as done by Einstein and Planck for electromagnetic radiation), Schrodinger introduced the quantum mechanical wavefunetion w. 2 2 In one dimension, this equation is given by: ~3~d Mx) + V(X)i,b(x): E§U(X) 1- What is the Born interpretation of that quantum mechanical wavefunction u}? 2m dx2 . ) ) ,. a k = {2 1' H I - I m 5‘ "i A: .r ‘1‘” (a; /72 en {J i fr; fife e; at? j Vi”- ? “t “’13 LL" . V _ J g , mi < '. Q; ’ L.“ a? jet/{1 we) My, FM 29 "I 17% / fl“? ’ f ' 32.; )5 #8311; Respond to the statement by TRUE or FALSE. Justify your respor/se in one of two sentences. 2— As a consequence of the Born interpretation and normalization of the particle wavefunction u}, u;(x) can be infinite. Vii-firgia’; “x ’ ~31 m If?) {gagiflé’ 7‘ a .3 3 i '4’? ’1 r I 9" . i A , 1 ,H y} if , f a raw: av e14 €54] lei/7 5;}, "1’ g i "’1’ ,5 a. V I 3- Because the Schrodinger equation imposes a second derivative on \J/(X), the wavefunction does not have to be continuous. i ’ 3,55,» i,/15J£ZWQ~ [x3 ,3“ a, M}; 172:; 3 an A. is 1 ’T‘éil/ég, . “1’”: W I I 54?) Li“ was; the. Gift? i7 flue“ ‘ 4- The wavefunction cannot be zero everywhere. 3 ._. . t " . , ., ,3 a 6,; iv i” "‘ > t” ‘ ~ a win m f“? 6/3/12 Uth t . v («be m); e y 1 max . .L. W i J {:5}? 1/7 1/? is?!“ W £52,? V3 Zig’fér ‘ l 5- w(x) must be square-integrable. ‘ {1/ X. J5 I ’ (ft 16/ i Extra {Points gamed here will be added 2‘0 your total grade) Exercise 2 (4 points) We assume that the state of a vibrating atom is described (in l dimension, x) by the 2 wavefunction 41(x) = A x x x exp[— x ] 2a2 The probability density associated with this function has a maximum value at a unique value of x : XM. 1) Write the differential equation that must be satisfied at x 2 XM. 2) Show that the most probable location of the particle is at x = XM = a. A 3" 1 3 7' v; w‘ ~ n- “" a a “ ’Z’“ 1"” ” “ Lil saga-i a”; a: new» m" (it Wm H” knee ,. . a” f f ’ J & flit/3m, e!» Z/M Xv 3e“? if) :14! m; i ){P’l ya: > i}; ; 3:1 M 7kg a"? Vfi’éWt/gfiffiiflfifl 7%;3/ X'M 2 1 1 . eat-[fig L a}: (Z;i W3?) ,5 A g, ‘ . _)i,/ 1.6:, 1’ “re (:11. 1:}: a)?!» L gaze mg; e at. .3 Fwy/4L 2“ 2.9: ‘ e '\ h ‘ 9L All}; :. "KQd’ Khx’t){i/ : ’3 5‘“) it: 1., 59,31 245,1 p” a ( {at (L [L ‘ '“ 7: (Bl Thad cl LN :: =24 Lib It ’ ...
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Phys chem II CHM4411 Exam 1 Solutions - CHM4411—02...

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