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1921_001 - QXWS‘WWQF Q9533 “— Hsmeoaqtfl*F—‘C W...

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Unformatted text preview: QXWS‘WWQF Q9533 “—- Hsmeoaqtfl *F—‘C W 2 2 6-12. E=—1-mv= ”h 2 = l 2 ZmL2 va 2 2 2mL2 (Equation6- -24) n2 [va )[zzhz )=[ ”h J 56-18. This is an infinite square well with L = 10cm. W“; he” 2 1 (2 0x10'3kg)(20nm/ y) E" :23le =2T‘2— 2(.3 (16x107sly)2 Rx I/am¢% ./ 3__ML 2' 8(20x10'3kg)(2(]><10"91ri1)(0.1m)2 n =______2____W. *._—- 2(3.16x107s) (6.63x10“34J-s)2 a42(2.0x10—3kg)i(20x10'9m )(0. 1m) (3.16x107s) 6.63x10'34j.3) a «38x10“‘) 1:: in __ 6—22. lyn(x)=\/%Slnn:x __ 7 7- ______ L _ . u, __ To Show that Isin£1mjjsin£nmedx = O 0 L L fi— Using the identity 2 Sin A Sin B = COS(A e B) _ COS(A + B), the integrand becomes ‘wm— l §{COS[(H ~171)7rx/L:l— GOSH}? + i31)7rx/L]} L Sln n—m izt/L The integral part of the first term is ___(__w_i;__m 12' (11‘ “- m) tenn with (n _+ m) replacing (n — m). Since :2 and m are integers and n i m, the Sines both and similarly for the second mh, vanish at the limits x = 0 and x = L. L . - '. ISin[E7—r£)sin[nmx)dr : 0 for n i m. — 0 L L L2 L2 I __ L. “3*er (See Problem 6-29.) And (x)_-i 2 2 2 112 112 ‘b 0.= (x2)—(x)2=[£__L_2_£—] = [—1— 12] =0.181L a. “3% _ 3 277: 4 12 27: 1mg 5 @ 7:th - (pi) = and (p) = 0 (See Problem 6-31) 2 ”2 0' = (p2)-(p)2 =[if 4] =1? And app=(o.181L)(nh/L)=o.563h 6-43. (a) Forx>0, hzkjlzmH/g =E=azkfilzm=2Vo 112 So, k2=(2mVD) /T1.Becausekl =(4mr/0)“2/h, then kzzkl/JE ‘ (b) R=(Ie—k2)2/(ki+kz)2 (Equation 6-68) Hawk m4“ 2 =(1—1N5) /(1+1/\/§)2 = 0.0294, or 2.94% ofthe incident particles are reflected. _ (c) T =1—R =1—0.0294=’0.971 «9((1) 97.1% of the particles, or 0.9718106 2 9.71 x 105, continue past the step in the +9: direction. Classically, 100% would continue on. ' mugs-=44. (a) For x > 0, 7121:: 12m — VD =: E = hzkf I2»: = 2V0 So. k2 =(6mVo)”2/h. Becausekl =(4mVo)m/h. then k2 =x/375k1 (b) R=_(I~a—k2)2/(Ia+k2)2 ' R-=(k. "k2)2/(k1+k2)2' =(1ufil3)2/(1+J3/—2)2 4.0102 01' 1.02% are reflected at x = 0. (c) T=1-—R =1—0.0102=0.99 (d) 99% of the particles, or 0.99 ><106i = 9.9 x105, continue in the +x direction. Classically, 100% would continue on. 6—45. 648. Using Equation 6-76, Tm16£[ (a) and aa = 0.611111 ><11.46nm‘l = 6.87 Since era is not << I, use Equation 6-75: L)- The transmitted fraction / i I = 2(0.5llx106eV/c2)(eV)/h 0.6 nm=a - C 2260‘” = 1 1.46m" = 197.3eV-nm sinhzaa - “i 81 . , 2 7 A T=[l+_—_m4(E/fi](l—E/n)] =[1+[§5]smh (6.87)] Recall that Sinhxri(e’——e"‘)/2, \+ +(4/QU‘A4)“ 48:9 <5) T =[ ”gr-“3 E - 1“ I ¢ u g . _ a (b) Noting that the Size of T13 controlled by am through the slnh2 am and increasmg T suits 81 4 = 8618‘ €6.87 .. €4.87 2 l + —86[————-—-—-—] = 4.3 x 104r is the transmitted fraction. implies increasing E. Trying a few valuesa selecting E = 4.5eV yields '2" --= 8.7 x104 “—— or approximately twice the value in part (a). ‘/ V l 0 2.0 6.5 1_£ V 0 l1“ e'2“" where E = 2.0eV, VO 2 6521/, and a = 0.5mm. a 6 JEW'WQ z 6.5 x 10-5 (Equation 6—75 yields T = 6.6 ><10-5 .) gm? WV’A-S’ ‘.. r... 6-49 lie—M andT-I—R (E ' . — 2 — quatlons 668 and 6370) (kl + k2) (a) For protons: = \/2nic2E/i-‘ic = .{2(938MeV)(40Mer/)I197.3MeV- fm = 1.388 1mm. fingL— f k2nJ2mc2(E—Vo)lhc=./2(938MeV)(lOMeI/)l197.3MeV-fiaz20.694 zmtéfi —--g R_[1.388—0.694] _[Q__.694 — 0.111 And T =1ag=l‘0.889_, 1.388+0.694 2 08 %_ (b) For electrons: L k =1.388[051I 938 0.51} w)”: =2 0.0324 k2 : 0. 694(— 938 ——”]z =0.0162 R_ [0.0324— 0.0162 =0.111 AndT:l—~R=0.889 m————c 00324+0 0162 , _ / —%— N0, the mass of the particle is not a factor. (We might have noticed that J; could __fi be canceled from each term.) 6—51. (a) The probability density for the ground state is P(x)=z// 2(x )= (2/ L) sin2 mt! L. The probabllity of finding the particle in the range 0 < x < [/2 Is: LIZ “Fifi “51/" ER. #12 = IP(x)dx=-— Em! Sir:2 1.1c171,;_=-—72[[——0)=l whereu=7rle 0 Lit 0 4 2 L13 ni3 . . L (b) P: IP(x)dx= 2L Sin Nudu:£[£—MJ=l—£=OJ95 0 L7: 7;“ 6 4 3 4n (Note: 1/3 is the classical result.) 3LI4 Bald "" (c) P: [PM 2” i sin udu=3[-3£ Sfl‘o’iliji-tiwsw . 0 L77: 7r 8 4 4 272‘ (Note: 3/4 is the classical result.) 7. T :16§[1—§-]e‘2‘“ whereE = IOeV, P; = ZSeV, and a =1nm. o a _ _ (a) a = 1/2m(Vo —E)/h =1/2(m0c2)(2; —E) hc _____ _" =1’2(0.511x106eV)(15eV)/197.3eVonm =19.84nm'1 m And aa = (19.84nm-‘)(1nm) = 19.84; 20m = 29.68 _ T a lav—OX1 431 82—29-53 a 4.95 x 10"13 25 25 (him a =70.1mm: aa 2 (19.84nm-‘)(0. 111m) = 1.984 T z 16[R][1—419]e'”63' a 0.197 25 25 ...
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