Exam_1_Crib - CHER/1370‘ . . . _ . ._ _ EXAMI r ' '...

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Unformatted text preview: CHER/1370‘ . . . _ . ._ _ EXAMI r ' ' Tuesday, February 5, 2008 I Name EXAM Q This exam consists of four (4) questions on ten (10) pages. Also included as the last pages are a Table of Constants and Conversion Factors and an Electromagnetic Spectrum. You may detach these pages if you'wish. Foifmulas are embedded in the questions for which they are needed. You must show all work and reasoning to obtain any credit. For. full'lcredit include the _ correct units in your final answer. l. (30 points) _ a. (10 pts) Recall for a complex number 2 = X + iy , the real part of z E Rez = x and the imaginary part of z E Imz = y. Also 2* E X miy and exp(i(|)) = cos¢ +isin¢. Finally recall that complex exponentials like exp (hp) obey mathematical rules identical to those obeyed by real exponentials like exp (¢). (i) (4 pts) If 2 =(x+iy)exp(i¢) find Re(z*). 2 : WM) ( Cashew) -..—.. Wqfi _ 75m?) +(w595 +«5ml)i 1242*) «.4 om; .. yin/if ‘ '(ii) (6 pts) Show that ii = exp[—£]. [Hint Find the (bifor which i=exp(i¢) .] 2 ’09 .33. 9. ._rL __ : “L ‘- ) ‘6 b. (10 pts) Determine whether the following operators 0 are linear or non-linear. To do this use the condition for a linear operator _ é[;°iwi (2)] = :CiO‘Pi (Z) I (1.1) E where the ei are complex constants. (i) (5 pts) O=[ ]“ where ow(z)ewt(z). a [G ‘2‘. (’X) «T crew] 2: [me m + (acme) x ' I z: 6.96%) + cf 7%) ((3 W7 t edits-M I j Cfifim‘) + a 9%) ¢ BEG‘MMHmWJ t c; Since) + (28%“) Neal/{91W (ii) (5 pts) 0 =[w—%:?+U(x)] where ow): —2—?:nmag:u(v5w(z)+U(x)ip(z). I 6FC!%CK?t&%W9] 1:— fimewfl + um [G‘fimflt Mug .fL n— 'L e em A A + (l Metre) _~_— 6.0%)er (-40%) ' Linew . c. (10 pts) In this problem recall that: 6054) 3 W p (1.2) and that (13(2) is linearly independent of w, and L112(Z) if (13(2) :2 clipI (Z)+C21|}2 (z) for all complex constants c1 and 02 . (1.3) 2 . (i) (4 pts) Consider the operator Show that exp(ikx) and exp (—ikx) are x . d2 .. _ linearly independent eigenfunctions of d—2 with common eigenvalue —l<2. Thus x the eigenvalue 48 is at least twofold degenerate. _ I MGM) : e—tzm’tilex) I % meme) :: - P: Whit“) M s; e'XtDC'fbx) Wt ewe-M) we eeamwcw t ti? W came/Ion WWIle elf . gemse expde 7E mam-1%) HAW) .W 0%) .3; mg WCflm _, ' se mating/<3 W3 indffmeéme tween, Owl Weill“) . Lem/m, 99 cm «a; 2 (ii 6 pts) The eigenvalue —k2 of i» is not three-fold degenerate even though dxz . 'J 2 with eigenvalue —k2. This is because coskx is cos kx is an eigenfunction of d x not linearly independent of exp(ikx) and exp (—ikx). Prove that 2 ficoskx = —k2 cos kx but that cos kx is not linearly independent of exp(iikx) . x , goal: 0553200 :: fifiéwlw) :— Glfiffiflm) :3. )4 fiWflxj a («WWW :—|a°‘0t:5ix . Dome rt 42m ‘ (MM '2 not (Mam melka at; Wffikp‘) ‘ 2. (25 points) The energy density u(v,T) of light of frequency v emitted by a bIackbcdy at Kelvin temperature T is given by Planck’s formula up] (v,T) = 811:?)3 [exp[:—;) — I] . (2.1) a. (2 pts) Show that in the high frequency limit, Planck’s formula reduces to an equation given earlier by Wien; namely WAVE): 8731""; “pH—fl. (2.2) That is, Show ‘ “1 lim P, v,T =uWi v,T . k had} ( 3 _ \ MM )T< ) “brew gem ALF; lim aim" 2:. a: elm”; - - was; if \{Aab 5 bit _. A gfihfiexT : [AL-£6513 Thee, . 819$) [extaLK—f-B ‘g : c3 —3 b. (18 pts) (i) (10 pts) Show from Eq- (2.2) that the frequency VMAX at which uWi (V,T) has its ' maximum value at temperature T is _ VMAX=£;1-T—. (2.4) ---?’ vmx, AU“; : O “T 2496 [— Aw \‘ lk-y junk)? efkvlk : O _. ‘9 a“ \3V3 C... )1- + C5 . AW" ‘- *2? ' ‘5 x \k’t Av \ 8mka 6.. v c‘a‘ 33°03 om biéeg, (ii) (3 pts) Recalling v = %, show that the wavelength of light lMAx corresponding to VMAX is given by he 7L T = — . 2.5 MAX ( ) . $8— vroow ' ‘1‘: A? c _ 5&3" {—9 “F” " P” fl? 1-mont- h .T , a c: ‘1 3c.» _ ' Arm”)! i’ (iii) (5 pts) Using the Table of Constants at the end of the exam, find the numerical ' value of £15: in mK units and enter as Eq. (2.6). 3k t? -1 h g, . e, sxvo‘i’ubs xl-q“ X‘, "‘5 ifs—a €13: a gigging-2533“ mK. _ (2.6) gigs” q.q.qx\o'3m\< c. (5 pts) Solve the following problem using your results in Eqs. (2.5) and Eqs. (2.6). In a hydrogen bomb explosion T can reach ~107K . What value of XMAX does this correspomi to? Give your result for AM” in A recalling 1A =10_'°m. What region of the electromagnetic spectrum does XMAX lie. in? _E> l 5‘): “a -ocmxw m“ 'mqk b Consider a molecule with three equally spaced energy levels 80 > 0,3[ =_ 230 , and 82 = 330 . The degeneracies of the levels are, respectively, g0 = Lg, = 2 , and g2 = 2 . Thus the molecule has five quantum states. In this roblem', consigr a gas of such molecules at temperature T. g 2 u. a __ _ 3. (20 points) EU): e M +26 KT + 2 e “T _ Determine from the Boltzmann distribution the probabilities P(sO,T),P(el,T) , and P (22$) [Recall the Boltzmann distribution P(8i ,T) = 2—] (T) exp (—1»:i /kT) g]. where Z(T) = Zexp(-8j/kT)gj .] allj a. (lOpts) as T—>ooK. if? “£461? + 2;? W ZCT): mm ((3 T+ 2e "1"” T9“ . I I " P(€blT) : ’3: 9“ P(£i].T)‘—:* Tia: 4?" new a b. (lOpts) a_sT—>0K. l c F(€~,T):(Q‘W\ a #23,: F29, 1' “MT-TEL? ~90 §W+2e ‘T -29” “oi-£3! L W ' sziwfl2¢M~ __F4_,._, #1:" P(£H ) (Ho €P%T—+Qe%+2’€ lot 'f—w [+26 tar-l2? Pf . “5&2 __ '3: . A: "T i 26 KT , 0 lDCafOzl/WA “g: 725.. 4% I “m eta» ,24— ‘" [+0443 w '6 rifle Via??? “F” [+26 Till—29 W. 4. (25 points) Consider a particle of mass m moving in a one-dimensional baox with range 0 S x S a . The time—independent Schroedinger equation for the particle inside the box is (recall it = h/21r) hZ d2 A _—2;dX—2Wn(x)=Enw:i At the box edges 0 and-a 11;“ (X) _obeys the boundary conditions w“ (0) = w“ (a) = 0. (4.2) We showed in class from Eq. (4.1) and (4.2) that ' 2 2 wfl(x)=J§sin[fl] and E“ = “ hz ,n=1,2...oo. (4.3) a a 8ma a. (10 pts) Show that 1y" is normalized. That is Show that a i " 2 nnx d m— ' 2 —— d 21. 4.4 "W" (X)U‘Jn (x) X a InSln [ a J X ( ) . . . . . 1— 2 To prove Eq. (4.4) use the standard trigonometric 1dent1ty 31112 y m w. 2 Q - ' A a” \HCEEhnx/GBA'BC i‘iSblq—gagklag L 2 - O D clog-t £01 Wax} 661%: C}: _2. a {EA} 0 b. (15 pts) (i) (8 pts) On the graph below sketch @141. (x) = sin and a éwz (x) = gimp—2E]. Ewlorflx) 7 (ii) (7 pts) The wave functions 1y,(x) and 1412(x) we orthogonal. That is I:W1*'(X)W2 (X)dx = ‘3" Sin [Ejsin [fljdx = 0. a a Show that Eq. (4.5) is true using your plot. DUB A13 gimme/*3; q) 0420 q 2 £1— Wi oseo Bake CV2 m U? was \oe‘m‘e q/Q.‘ «Ale/0‘2 Q L “(233 41 .. 0 “La _ SK” YT: ' & CA 10 (4.5) 11 . 333$ ._ U” _. m. _ field Grange “6* .— ... 89” r G 8. .m l'ue , m8. 3% let i W , ‘ N? 3+ 3 “5 my f, $fi 759m Atmsar‘f'fiflim’ Chev-117mb; 5‘“ ad. (19%).12; 5451- 12 i ' values Of constants and conversion factors Speed of fight in vacuum at = 2.99792458x 101“ ms“‘(defined) 7 '- P‘lanc'k's constant I: = 6.62608‘21073‘ Is, a '= 3:127: = 1‘054573'x10‘3‘ Js. Pmron Charge . . +eg+1j6021779£1049 C, ‘ ' ' Electmn charge ' —e = al.6021-71x1'o"? c‘ ' Baltzmann’s copslam 1.38066mg"?3 '1 K7] _ Avogadrd’sr numbeg NA = (3.02214le13 me ‘1 Ré'st‘ma-ssrdf elactiron lm, : 9.10939x10‘2“ kg Res} mass of pfoton mp =1.67‘2623x10‘” kg 1 Rest mass of neutron m". '= 1.614929x10‘27 kg ' Gas cénsgant R = N-Ak = 3.3145 Iago!" K" . Faraday’s constant .F = 96485.3 .Cmolf' Rydberg cpnstant for H: RH -—- 1096716, cm" Bohr radius - r 7' £10,: 52.91772 pm .Bphrinagncton ' 3, =9.2740,2-x?10““_ J'T" Nueiea‘r magneton 16'” = 5.05079x10’z77 3 TI Permittivity of- vacuum £0 = 8.8’541878'x10‘” C2 J‘I m“ VEIBCtanjg value 3‘, = 23023193044 Protfin g valué 7_ ‘ g, = 5.585695 Pi (matheméticél). - 'n = 114159265359 Metfic prefixes: ‘ 7 _ Symbol ' Prefix Factor ' Sy'mball Prefix Factor 1d - . deci 0.1' da ._ deka 10 c ce_n_ti 0.01 _h . 'hemo 7 169- m mill-i 13.001 k - kilo ‘ ,9 micro ' 10-6 I M- mega 10.“ n . name .10" G ' giga 10“ p 'piccr ' 19"” I '. tera 19" f femto 10'” P pad I 1935 a {mo ‘ 10-“ ' ' E gm" ' in" "Energies: ' l-erg 2107' J '= '10-7 kgm’s" ' _ 1 eV :71.5o'21'z'7x10"" J - ' 1ca1=.4;134.1 7 ‘ .1 cm" 21.98645xm43 J Misc. ' '1}; =10"° m =10“’cm ‘ ‘ 7 Jim ="1Nx‘n" =1 kgm"s"2 ' .1 am =71 gmol“ ='1-.66054x10*=‘ kg ' uni/h = 0.447D4ms“ ...
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This note was uploaded on 06/27/2008 for the course CHEM 370 taught by Professor Oldsleepyman during the Fall '08 term at Purdue University-West Lafayette.

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Exam_1_Crib - CHER/1370‘ . . . _ . ._ _ EXAMI r ' '...

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