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Homework4 - :P_c?fm>-a3 1,3222 23.33.2223 324.333 2 16...

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Unformatted text preview: ____:P_c??fm>-a3 1,3222..- 23.33.2223 324.333. __ _ _ 2- 16 4He-)3H+p+e ’ _ .. __ Q: [nz(3H)+mp +m —m(4He)]c 2 AE AEu_ 4. 2eV c2 uc2=931.5x106eV (a) Am— - _9 . (b) error: ___‘Am—_~—_— —i§x—10—lm7.7x10'” =7.7x10f9 % m(Na)+m(Cl) 231: +35 Su , 2-20 WW -—..—> o :3 o--+v .. -- - - M 1.01M (a) before photon absorbed: after photon absorbed: — E = hf +Mc2 , E =Efinefic +1.01Mc2 Conservation of energy requires: hf + Mcz— - Ekineflc +1.01M’c2 kinetic hf=0.01fl/Ic71 +E,d . >0.01Mc2 mam: Real-ranging, the photon energy is: hf =1.01Mc2 __ Mcz + E greater than 0) must be supplied by the-photon. 6 ---——--—- 2-22. AB = Ame2 Am = Alf/CZ = M = 3.577x10‘25g (5.61x10329V/g) 223. (a) E 3 =—kT peratom ,,,,,,,,,,,,,,,,,,,,,,,, _ 2 _ 3 —23 _ __ _._ fl —§(1.38x10 J/K)(1.50x107K) W E=3.105><10"6J/at0m H atoms/kg = “cg/1.67 X It)"27 kg / atom = 6.0x1026a10ms Thermal energy/kg = (3.1 x lO'”J/atom)(6.0><IOZGatoms) = 1.86x10”J (b) E=mc2 :> m=E/c2 m=tRfixm“,r/p2tomymrfiba = 2809. 450MeV+938. 280MeV+0. SllMeV 3728. 424MeV= 19. 827MeV " " " ___. u=4,5x10'9u‘ W, ._ .. n .-_.___,_W (b) The photon’s energy is hf > 0.01Mc2 because the particle recoils fiom the absorption” 7. of the photon due to conservation of momentum. The recoil kinetic energy (which is ,2, _ ‘ _ ‘ _ _ 2-26. E 2 = ( pcz )2 + (me2 )2 (Equation 2-31) we):wmafia-:21“ =mc2[1+(p2/m2c2)]mzmc2[l+%( /mc)2+~-]=mcz[l+ pz 2] 2mc =mc2+pzl2m ' 2-27. E2-m(pcl)2+(mc2)2 (Equation 2—31) (a) (pc)2 = E2 —(mc2 )2 = (5114!.qu2 -—(O.511MeV)2 = 24.74 or, p = #24374 /c = 4.97MeV/c (1))E=7’mc2 —> }'=E/m62=1/~Jl—u2/c2 ——> 1~—u_2/_c2_=(mc2/E)2.. .i. ”2 we =[1—(mc2 {EYTZ =[1—(0.511/5.0)2] =0.995 .7 2-29. 7E2 =(pc2)2+(m2)2 (Equation 2-31). 0746114er =(500MeV)2 +(mc2)2 _ mc2 =[(1746MeV)2 —(500MeV)2:l”2 =1673MeV —> m 21673MeV/c2 Ezymcz ~—> y =llm=Elmc2 m 1.12 u/c=[1—(mc2/E)2] =[1—(1673MeV/1746MeI/f] 20.286 —> u=0.286c' ' " 3-3. 8 u. pc 2 2 2 - B=—, u=—-,and c: E —— me u c E p ( ) pc 2 ‘/(0.561Me V):l ——(0.511MeV)2 = 0.2315MeV E: 0.2315Me?’ 20.41 c 0.561MeV _ 2.0x105V/m ' B 21.63x10'3T2163G 0.410 3—6. (a) 1/21fiu2,=Ek,sou= (2Ek/e)(e/m) u = [(2)(2000eV/ e)[1.76 x10110/kg)]”2 = 2.65 x107m'ls *V_ _ — (b) _ At =i:__9£.5.fl_._.=1,89X10—93=1.89ns _ _ __ _ Wm “___ ' i 'u 2.65x107m/s ‘ ' "___—___”..- (c) may =le zeEAti r 7777777777777 7 uy=(e/m)EAt,=(1._76x10”C/kg)(3.33x103V/m)_(1.89x10’9s)=1.11x10"m/s 7 ,fi,‘ 3-12. ,1”; = 2.898x10'3m-K i ___ _3 ‘ (a) 1m = 2‘89'8“_—"—x3112 m K = 9-66X 10—4111 = 0.966mm ,,A,W,,,,,,i ___...” ......... , ...... ,.,,_. - ' _3 I . ._.._.._... .._._ W. ‘. , ,"fl”Wi,A.,g,,,,,,,fifi,fi,w,”MM.-.“ (b) in. = 3W = 9.66x10‘6m = 9.66pm -3 ' -' - __ _ __ _ (C) ,1,” = “___—2.893531%}: K = 9.66x10‘7m = 966nm - 313 Eduéiidhé-4:h=&r4. Equatmm; Rzicy, * From Example 3-4: U = (8x5k4r4 )/(15h302) 4 __n_._.-_.__._..__,...._____..._.___ 2735 (1.38 X ] Or—23J/K) .. . . , , ,WAVWWWWH, :_”—“_—“———3=5.67x10'8W/m21(4 ___ , 15(6.63><10"34J-s)3(3,00x103m/3) r yr , _ hC/zl. hc/(thc/kT) 0.11:1” H 3'18. (a) Equatlon 3-17: E=W=W=mzfl951kT _ hcm ._ hc/(0.1hc/kT) _10kT (b) E=W"WS— ° I =4.59x10‘4kT e e —l Bquipartition theorem predicts E = AT. The iong wavelength value is very close to . kT, but the short Wavelength value is much smaller than the classical prediction. 325. (a) hf = Ital/l = 0.47eV. he (4.14x10"‘5eV-s)(3.00$<103m/s) = 2.55x10'7m = 255nm ,1 =‘ _ ”'3“ 4.87eV 4.87617 It is the fraction of the total solar power with wavelengths less than 255nm; i-e-, the 5- area under the Planck curve (Figure 3-6) up to 255nm divided by the total area The latter is: 12:01“ =(5.67x10'8W/m2 -K4)(5800K_)‘ =6.42x107W/m2. Approximating the former with um M with ;t z 127nm and AA, = 255nm: 87rh¢(127x170'9m)_5 - '[“(127"m)](25_5"m)= e"°”"l‘”"'°")_1 (255x109m)=1.23x104J/m3 R—(0‘255nm)=5(1.23x10*‘J/m3) '_> W 4 R 3.00x108m/s 1.23X104J/m3 _ , ( )( ) —4 2W M fiamzlmo = 653nm, j; =%-——3'_99_V—_ = 4.59x10‘Hz £13 __ 1240eV-nm 4.136x10"”eV-s gal l .Qev , 1 he 1 124deV-nm 0’) o eh ¢] ii 300nm 9) ' 3-26. (a) ,1, = hc J .1[1240eV-nm 1. (C) VB _Z[_:t“¢ 400m —1.-9eV) =1.20V e 3-30. Using Equation 3421, ' 5_nW._.___________..m--.W--.i#,--.--______- .. Solving for 12 yields: .11 = 6.56 x 10-3418 . Substituting 1: into either (1)6r (2) and s. Iving . h c {I}. I 1 035 = _ ___.___________ ___. . ....W W--. ___ W,WA__ .. ( ) e[_435.8><10'9m] e h c 65 2 0.38 =— ———-————— —— ( ) e[5461x10'9m] e ' ___W W - . he 1 1 Subtractm 2 fi'om l , 0.57: ___—___... g( ) ( ) “lo-{435.8 546.1] for We yields: ¢le=1.87eV. Threshold frequency 1S given by hf / e: 65/ e or 1.87eV 1.60x10“9c f:[£][£]:£—————)—(———_3?-——~Hl= 4.57X10l4HZ h 6.56x10 J-s 60 663 10'34J-s 3.00 108 I E— nfE=m____:_____L__‘_____( )( X )gm x m S)—217><1017J 3 550x10 _ hc P1240eV'onm A. ' 653nm . ’16 1240 V. ...-........_.._...._ . . . WW.W.WWMW.. 3...__.___..____ ___ =-——¢=—~—5———’1”—'—1 .=90eV 223eV l 300 Onm ...
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