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原子物理学第四版-课后答案---标准版

原子物理习题库及解答第一章1-1 由能量、动量守恒 ⎪⎩⎪⎨⎧'+'='+'=e e e e v m v m v m v m v m v m αααααααα222212121(这样得出的是电子所能得到的最大动量,严格求解应用矢量式子)Δ得碰撞后电子的速度 e em m v m v +='ααα2 p故 αv v e2≈' 由)(105.24001~22~~~4rad m m v m v m v m v m pp tg e e e e -⨯=='∆ααααααθθ1-2 (1) )(8.225244.127922fm ctg a b =⨯⨯⨯==θ (2) 52321321063.91971002.63.19]108.22[14.3--⨯=⨯⨯⨯⨯⨯==nt b NdN π1-3 Au 核: )(6.505.4244.1794422fm v m Ze r m =⨯⨯⨯==αα Li 核:)(92.15.4244.134422fm v m Ze r m =⨯⨯⨯==αα1-4 (1))(3.16744.1791221Mev r e Z Z E m p =⨯⨯==(2))(68.4444.1131221Mev r e Z Z E mp =⨯⨯==1-5 2sin /)4(2sin /)4(420222142221θρθr ds t A N E e Z Z ntd E e Z Z N dN p p ⋅=Ω=42323213)5.0(1105.1105.11971002.6)41044.179(⨯⨯⨯⨯⨯⨯⨯⨯=--68221090.8197105.144.1795.102.6--⨯=⨯⨯⨯⨯⨯=1-6 ο60=θ时,232221⋅==a ctg ab θ ο90=θ时,12222⨯==a ctg a b θ 3)21()23(22222121===∴b bdN dN ππ1-7 由32104-⨯=nt b π,得nt b π32104-⨯=由22θctg a b =,得 23233232)67.5(1021811002.614.310410104)2(⨯⨯⨯⨯⨯⨯=⨯=---οntctg a π )(1096.5224cm -⨯=)(8.23161096.5)41(2sin )4(2442b a d d =⨯⨯⨯==Ω∴-θσ1-8(1)设碰撞前m 1的速度为v 1,动量为p 1。

碰撞后m 1的动量为1p ',m 2的动量为2p ' 由动量、能量守恒可得:1211011p m m m n v p ϖϖϖ++='μ1212012p m m m n v p ϖϖϖ++-='μ其中2121m m m m +=μ,将它代入上两式可得:1211021121p m m m n m m p m p ϖϖϖ+++='1212021122p m m m n m m p m p ϖϖϖ+++-='它们之间的矢量关系可用下图表示,其中圆心C 为质心,0n ϖ表示质心系里m 1碰撞后的速度。

211121121,m m p m AO m m p m v OB OC +=+===μ当 21m m <时,A 点在圆上 21m m =时,A 点在圆上21m m >时,A 点在圆外由图可知, 12max sin m m AOC O L ='=θ(2)因ο90,1sin ,max max 12=∴=∴=L L m m θθ(请参阅朗道的力学)1-9 对Au 核:)(114144.17912211fm E e Z Z a p =⨯⨯==对Ag 核:)(7.67144.14712212fm E e Z Z a p=⨯⨯==由22θctg a b =可求得 )(12673.327.67)(21373.3211421fm b fm b =⨯==⨯=%30%70222121⨯+⨯=∴t n b t n b NdN ππ3331082.51025.11057.4---⨯=⨯+⨯=(其中 3231105.11971002.6-⨯⨯⨯=t n ; 3232105.11081002.6-⨯⨯⨯=t n )1-10 ⎰=∆2sin sin 2)4(42221θθθπd Nnt E e Z Z N ⎰=2sin sin )4(242221θθθπd E e Z Z Nntb a E e Z Z nt N ]2sin 21[4)4(222221θπ--⨯⨯=(1)94121041.1242.01024.61038.9⨯=⨯⨯⨯⨯=∆-N(2)104121076.131024.61038.9⨯=⨯⨯⨯⨯=∆-N(3)114121068.71311024.61038.9)10(⨯=⨯⨯⨯⨯=≥∆-οθN1211121061.81068.71038.9)10(⨯=⨯-⨯=<∆∴οθN第二章2-1 (1)3301053.69.1104.12⨯=⨯==E hc λ(Å))(1059.4101053.63141030Hz C⨯=⨯⨯==-λυ(2)3301065.35.19.1104.12⨯=+⨯=+=eE E hC λ(Å)2-2 利用公式 2202222,,n RhC Z E C n Z v Z a n e Zm n r nn e n -====αη 1311121104.12,V eV hc e E E V ⨯==-=λ (1) H 原子: 529.0221==em r e η(Å),)/(1019.261s m c v ⨯==α 12.2441222===r em r e η(Å),)/(1009.12162s m c v ⨯==α He +离子:265.021221==em re η(Å),)/(1038.4261s m c v ⨯==α 06.1412==r r (Å),)/(1019.262s m C v ⨯==α Li ++离子:176.0529.0311=⨯=r (Å),)/(1057.6361s m c v ⨯==α 704.0412==r r (Å),)/(1029.32362s m C v ⨯==α(2) H 原子:)(6.131eV Rhc E -=-=He +离子:)(4.546.13421eV Rhc Z E -=⨯-=-=Li ++离子:)(4.1226.13921eV Rhc Z E -=⨯-=-=(3) H 原子:)(2.10)6.1340.3()(121V eeE E V =+-=-=12162.10104.12311=⨯==eV hc λ(Å) He +离子: )(8.402.1041V V =⨯=3048.40104.1231=⨯=λ(Å) Li ++离子: )(8.912.1091V V =⨯=1358.91104.1231=⨯=λ(Å)2-3 )(8.91)4.36.13(912eV E E E =-⨯=-=∆2-4 )(2.1012eV E E E =-=∆由能量、动量守恒可得质子的阈能:)(4.202.1022121eV E E m m m E th =⨯=∆=∆+=)/(1025.624s m mE v th⨯==∴2-5 (1)175403293/1018.12931062.8/)6.1340.3(1104442855--⨯-⨯⨯+--⨯=⨯=⨯==-e e e N N n现175141,1e N N n =∴= 故14932311093.0104.221002.6⨯=⨯⨯⨯=-N V (米3)(2)室温下氢原子)(09.126.1351.1,113eV E E E n =+-=-=∆∴=2-6 只观察到赖曼系的头四条谱线 1216 Å,1026 Å,973 Å,950 Å2-7 Rhc Z hc E E hc Rhc Z hc E E hc 22322121536,43=-==-=λλ RhcZ hc hc 21215)20108(-=-∴λλ413376.1315104.1288)(15883122=⨯⨯⨯⨯=-=∴λλRhc hcZ故2=Z2-8 利用c mv eV W h mv mv W h αυυ221)(2.276.138.40212122==-=-=∴+= )/(1010.326s m c v ⨯==∴α2-9 利用折合质量 2,22121∞='=+=R R m m m m m e μ(1))(06.12122οηA a er ===μ(2)V 电离=13.6/2=6.8(V) )(1.522.101V V == (3)243021215341=⨯==Rλ(Å)2-10e pp m m m m m 186=+=μμμ(1)311108.2186-⨯==a r (Å)(2))(25301861eV Rhc E -=-= (3)90.4104.1213min=-⨯=-∞E E λ(Å) 2-11999728.0)1()1(=++HDM mm m, 将50020.0=DHM m 代入 HH M mM m 50020.01999728.0)1(+=⨯+41072.2999728.01499528.0-⨯=-=HM m 310835.1⨯=∴m MH2-12 (1))/(26.31031067.1106.12.10,102719s m mch v C h mv =⨯⨯⨯⨯⨯==∴=--νν(2)反冲能 962221044.51093822.102,2)(-⨯=⨯⨯==∴=mch h E mc h E RR ννν2-13 利用选择定则 1±=∆l ,共有6条。

2-14 (1))(10697.1589311633-⨯==-m T T p s )(10447.240861163-⨯==m T p 上两式相加得,)(10144.4163-⨯=m T sJ eV hc T E s s 193310225.8)(14.5-⨯-=-=-=∴J eV hc T E p p 193310848.4)(03.3-⨯-=-=-=(2))(11.2)(14.53313V eE E V V eE E V sp s=-==-=∞第三章3-1 )(1039.12.11079.522245eV B B E B sz --⨯=⨯⨯⨯===∆μμ3-2 54511)1(2)1()1()1(1=-=++++-++=j j s s l l j j gB B B j j j g μμμμ55.11552)1(-=-=+= B B j jzgm μμμ)56,52,52,56(--==3-3 0)123(232)125(25)14(4)123(231)1(2)1()1()1(1=++++-++=++++-++=j j s s l l j j g0=∴j μ3-4 25.010.021027.9002.0)400(1066.187.1072242272⨯⨯⨯⨯⨯⨯⨯⨯--=⋅⋅=D d ZMv dZdB sz μ= 124 (T/m) (g J = 2, m J = ±1/2)3-5 5215242)1(2)1()1()1(1=-=++++-++=j j s s l l j j g321050253.01.02352222-⨯⨯⨯⨯⨯⨯⨯=⋅∂∂⨯==∆B ZMv Dd ZB Z Z μμ)(4.10mm =3-6 23=j Θ,且34=g ,故分裂成四条。

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