原子物理学 杨福家 第四版(完整版)课后答案
原子物理习
题
快递公司问题件快递公司问题件货款处理关于圆的周长面积重点题型关于解方程组的题及答案关于南海问题
库及解答
第一章
111,222,,mvmvmv,,,,,,,ee222,1-1 由能量、动量守恒
,,,mvmvmv,,,,,,ee,
(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)
Δp θ
mv2,,,得碰撞后电子的速度 p v,em,m,e
,故 v,2ve,
2m,p1,mv2mv4,e,eee由 tg,~,~~,~,2.5,10(rad)mvmv,,,,pm400,
a79,2,1.44,1-2 (1) b,ctg,,22.8(fm)222,5
236.02,102,132,5dN(2) ,,bnt,3.14,[22.8,10],19.3,,9.63,10N197
24Ze4,79,1.441-3 Au核: r,,,50.6(fm)m22,4.5mv,,
24Ze4,3,1.44Li核: r,,,1.92(fm)m22,4.5mv,,
2ZZe1,79,1.4412E,,,16.3(Mev)1-4 (1) pr7m
2ZZe1,13,1.4412E,,,4.68(Mev)(2) pr4m
22NZZeZZeds,,242401212dN1-5 ()ntd/sin()t/sin,,,,,2N4E24EAr2pp
1323,79,1.44,106.02,101.5123,,(),,1.5,10,, 24419710(0.5)
,822,610 ,6.02,1.5,79,1.44,1.5,,8.90,10197
3aa,,1-6 时, b,ctg,,,,6012222
aa,,时, b,ctg,,1,,902222
32()2,dNb112 ?,,,32dN1,b222()2
,32,324,101-7 由,得 b,bnt,4,10,,nt
a,由,得 b,ctg22
,3,3a24,104,10(),, 2,23106.02,10,ntctg2,323.14,,2,10,(5.67)181
,242 ,5.96,10(cm)
a2(),d1,244?,,(),5.96,10,16,23.8(b) ,4d,4sin2
1-8(1)设碰撞前m的速度为v,动量为p。 111
,, 碰撞后m的动量为,m的动量为 pp1212
由动量、能量守恒可得:
m,,,1,p,,vn,p1101 m,m12
m,,,2,p,,,vn,p2101 m,m12
mm12其中,将它代入上两式可得: ,,m,m12
mpm,,,211, p,n,p101m,mm,m1212
mpm,,,212,p,,n,p 201m,mm,m1212
,它们之间的矢量关系可用下图
表
关于同志近三年现实表现材料材料类招标技术评分表图表与交易pdf视力表打印pdf用图表说话 pdf
示,其中圆心C为质心,表示质心系里m碰撞n10
mpmp2111OC,OB,,v,,AO,后的速度。 1m,mm,m1212
当 时,A点在圆上 m,m12
时,A点在圆上 m,m12
时,A点在圆外 m,m12
,OCm2由图可知, sin,,,maxLm1AO
,(2)因(请参阅朗道的力学) m,m,?sin,,1,?,,9021LmaxLmax
2ZZe1,79,1.44121-9 对Au核: a,,,114(fm)1E1p
2ZZe1,47,1.4412 对Ag核: a,,,67.7(fm)2E1p
114b,,3.73,213(fm)1a,2由可求得 b,ctg2267.7b,,3.73,126(fm)22
22dN?,,bnt,70%,,bnt,30% 1122N
,3,3,3,4.57,10,1.25,10,5.82,10
23236.02,106.02,10,3,3(其中 ; ) nt,,1.5,10nt,,1.5,1012197108
22,,,,,ZZe2sindsinZZed221212,,()NNnt,1-10 ,2()Nnt,4,4,4,Esin4Esin22
2ZZe12,2b12, N2nt()4[sin] ,,,,,a24E2
12,49,N,9.38,10,6.24,10,0.242,1.41,10(1) 12,410,N,9.38,10,6.24,10,3,1.76,10(2) ,12,411(3) ,N(,,10),9.38,10,6.24,10,131,7.68,10,121112 ?,N(,,10),9.38,10,7.68,10,8.61,10
第二章
3hc312.4,102-1 (1)(Å) ,,,,6.53,1001.9E0
C3,,, 3,10140,6.53,10,10,4.59,10(Hz)
33hC12.4,10(2)(Å) ,,,,3.65,10EE,1.9,1.5e0
2222nanZ,0,ZRhCr,v,C,E,,,,2-2 利用公式 2nnn2nZnZmee
3E,Ehc12.4,1021 ,V,,,,11eeVV11
2,6(1) H原子: (Å), r,,0.529v,,c,2.19,10(m/s)112mee241,6 (Å), r,,4r,2.12v,,c,1.09,10(m/s)21222mee21,+6 He离子:(Å), r,,0.265v,2,c,4.38,10(m/s)1122mee
6 (Å), r,4r,1.06v,,C,2.19,10(m/s)221
1++6 Li离子:r,,0.529,0.176(Å), v,3,c,6.57,10(m/s)113
36 (Å), r,4r,0.704v,,C,3.29,10(m/s)2212(2) H原子: E,,Rhc,,13.6(eV)1
2+E,,ZRhc,,4,13.6,,54.4(eV) He离子: 1
2++Li离子: E,,ZRhc,,9,13.6,,122.4(eV)1
(E,E)(,3.40,13.6)21(3) H原子: V,,,10.2(V)1ee
3hc12.4,10 (Å) ,,,,12161eV10.21
+ He离子: V,4,10.2,40.8(V)1
312.4,10 (Å) ,,,304140.8
++Li离子: V,9,10.2,91.8(V)1
312.4,10,,,135(Å) 191.8
2-3 ,E,E,E,9,(13.6,3.4),91.8(eV)21
2-4 ,E,E,E,10.2(eV)21
由能量、动量守恒可得质子的阈能:
m,m12 E,,E,2,E,2,10.2,20.4(eV)thm1
2E4th?v,,6.25,10(m/s) m
,55N,(,3.40,13.6)/8.62,10,293,1.18,10/293,403,1758n,e,4,e,4,e,4,102-5 (1) N21
1175N1,Ne 现 ,?,n14
N,314931,,22.4,10,0.93,10V 故(米) 236.02,10
(2)室温下氢原子 n,1,?,E,E,E,,1.51,13.6,12.09(eV)31
2-6 只观察到赖曼系的头四条谱线 1216 Å,1026 Å,973 Å,950 Å
hc3hchc36hc2-7 ,,,,, ,,2212E,EE,E4ZRhc5ZRhc2132
(108hc,20hc) ?,,,,22115ZRhc
32hc8888,12.4,10Z ?,,,4Rhc,,15(,)15,13.6,133721
故 Z,2
111222-8 利用 h,,W,mv?mv,h,,W,40.8,13.6,27.2(eV),mv2,c222
6 ?v,2,c,3.10,10(m/s)
mmmR12e,,2-9 利用折合质量 , ,,,R,22m,m12
2,,(1) 21.06()r,,a,A12,e
10.2(2)V=13.6/2=6.8(V) V,,5.1(V)电离12
4(3)(Å) ,,,1215,2,243013R
mm,p,,,186m2-10 em,m,p
a,31(1)(Å) r,,2.8,101186
(2) E,,186Rhc,,2530(eV)1
312.4,10(3)(Å) ,,,,4.90minEE,,1
m(1,)mmHD2-11 , 将代入 ,0.50020,0.999728mM(1,)DMH
mm(1,),0.999728,1,0.50020 MMHH
mM3,4H?,1.835,100.499528,1,0.999728,2.72,10 mMH
,19hh,,10.2,1.6,10mv,,?v,,,3.26(m/s)2-12 (1) ,27101.67,10,3,10Cmc
2(h)h,,E10.2RE,,?,, (2)反冲能 6,9R22h,2,938,10,5.44,102mc2mc
2-13 利用选择定则 ,共有6条。 ,l,,1
16,12-14 (1) T,T,,1.697,10(m)sp335893
16,1 T,,2.447,10(m)p34086
6,1上两式相加得, T,4.144,10(m)s3
19, ?E,,Thc,,5.14(eV),,8.225,10J3s3s
19, E,,Thc,,3.03(eV),,4.848,10J3p3p
E,EE,E3p3s,3s(2) V,,5.14(V)V,,2.11(V)1ee
第三章
,,543-1 ,E,2,B,2,B,2,5.79,10,1.2,1.39,10(eV)szB
j(j,1),l(l,1),s(s,1)14g,1,,1,,3-2 2j(j,1)55
2,,gj(j,1),,,15,,,1.55, jBBB5
6226,,gm,,(,,,,,), jzjBB5555
3355(,1),4(4,1),(,1)j(j,1),l(l,1),s(s,1)22223-3 g,1,,1,,0332j(j,1)2(,1)22?,,0 j
2,272dB107.87,1.66,10,(400),0.002MvZ,,,243-4 9.27,10,2,0.10,0.252,,d,DszdZ
= 124 (T/m) (g = 2, m = ?1/2) JJ
j(j,1),l(l,1),s(s,1)242g,1,,2,,3-5 2j(j,1)155
232,,,0.1,0.3,5,B,B52D,Z,2Z,2,d,,, 2,3ZMv2,50,10,Z
,10.4(mm)
343-6 ?j,,且,故分裂成四条。 g,23
412,,,,,gmu,, jzjjBBB323
2,B22,3,Z,,?,Z,,Z,,0.60,,0.40(cm) ,,ZB33
3,312.4,10,29.6,U,hc,,,,3.67,103-7 810
4242Z1,(Z)mc,2220UmC,,,,, 又 302nl(l1),162
4++16,,U?Z,,81?Z,3 是Li. 2,13.6
4242(),,zmC,mC00,,,U3-8 3322(1),nll
42,U,mC0,,U,2B,?B,,,0.391(T) 又 B,32,2,B,2B
3-9 (略)
3-10 能级图:3S分裂成三条,g = 2 1
3p不分裂 0
,E,g,B,2,B BB
,5,E2,5.79,102B,,1~B,v,,,,93.4(m) 3,10hChc12.4,10,10
~ 故不是正常塞曼效应。 ,v,L
3-11 参照
书
关于书的成语关于读书的排比句社区图书漂流公约怎么写关于读书的小报汉书pdf
图15.7
,5B,5.79,10,2.5,1BL,,,1.17(cm) 3,8hc12.4,10,10
42,1,1由此可计算得 L,0.78(cm)L,1.56(cm)33
15,1,1 L,0.39(cm)L,1.95(cm)33
,1 L,1.17(cm)
424:3-12 (1) p,E,g,B,,B3BB23
224: p,E,g,B,,B1BB232 4s:,E,g,B,2,B1BB2
1,E,,E,2,B,,B,1.5,E (2) 21BB13
,,hc,3 而 ,E,,7.36,10(eV)1,,12
,33,0.5,E3,0.5,7.36,101?B,,,27.2(T) ,5,77,5.79,10B
,,e,,(gSgL),,3-13 (1) sl2m
e,BU,(2m,m)(2) sl2me
,E,2,B(3)3S态的能级分裂: B
,E,,B 3P态的能级分裂: B
能级图见p图书馆5.11。 115
B,~Bv,,3-14 hc
5,5.79,10,422~(Å) ?,,,,,v,(1210),,0.0273312.4,10谱线分裂为三条: (Å) ,,,,,1210,0.02730
(Å) ,,12100
(Å) ,,,,,1210,0.02730
第四章
4-1 ,E,E,E,0,(24.5),24.5(eV)1,1
2,E,E,E,0,ZRhc,4,13.6,54.4(eV) ,22
,E,,E,,E,24.5,54.4,78.9(eV)12
,,1122222L,S,(J,S,L),,L(L,1),,,3,4-2 22
,,51,2LSLS[J(J1)S(S1)L(L1)],,,,,,,,,,,,,,22J4-3 , ,,,3132,LSLS[J(J1)S(S1)L(L1)],,,,,,,,,,,,,,222,
2224-4 S,L,J,2LJcos,
222(J,L,S) ?cos,,,0.94262JL
,, ?,,1930
?4-5 价电子数为偶数的氦,Be, Mg, Ca,可能出现正常塞曼效应。S可能为0。
12,4-6 l,l,s,s,,先算L-S耦合 12122
,43210(,0时)5种可能状态SJ,,,,
,
,1,54321(S,1时),,L,,,,L,4,3,2,1,0,,, ,,L,4,3,2,1共18种,,S,1,0J,,,L,1,321013种可能状态,,,
1515,,,,l,,l21,,,,2222j-j耦合 ,j,j,,211313,,l,,l,,12,,2222,,
55,jj,J,,,,,,,,时,54321012,22,53,jj,J,,,,,,时,432112,22共18种态,且出现相同J的次数也和L-S耦合相同。 ,35,jj,J,,,,,,时,432112,22,33,j,,j,时,3210,J,,,1222,
424-7 (1)np形成的电子态与np相同
S,1S,0L
31DD3,2,122
13 1PP12,1,0
130SS01
1313如考虑泡里原理只有,其中能量最低。 DPSP22,1,002
22251P,PP(2)np形成的电子态同np相同,故只有,其中态能量最低。 313222
33 (3)同上题的LS耦合,由于非同科电子,故有,其中能量最低。 GG5,4,33
4-8 (2S,3P)所形成的原子态为
L S = 0 S = 1
31 P P 12,1,0
根据跃迁的选择定则 ,s,0,,l,,1,,j,0,,1, 共可产生10条光谱线:
(若该电子被激发到2P态,则只发一条光谱)
1 故J = 0,必定有的基态。 4-9 ??m,0?l,0S,l0l
?m,0?s,0,ss
4-10 m 2 1 0 -1 -2 m ll12
4 3 2 1 0 2 m,mll12
3 2 1 0 -1 1
2 1 0 -1 -2 0
1 0 -1 -2 -3 -1
0 -1 -2 -3 -4 -2
S = 0时,M= 4 3 2 1 0 –1 –2 –3 –4 L = 4 L
= 2 1 0 –1 –2 L = 2
= 0 L = 0
S = 1时, 对角线上值不能取
? M= 3 2 1 0 –1 –2 –3 L = 3 L
= 1 0 –1 L = 1
313131FG,F,D,P,S?有其中最低。 44,3,222,1,002
1S,J,0,,,0?4-11 基态氦原子是,故不分裂,只有1束。 0J
12 而硼原子的基态是 ,分裂为2束。 PJ,122
3234-12 的电子组态为 L,m,1,0,1,0PS3P,l15l
1113,,,,, Sm,s2222s
3J,S, 2
4 故基态为:S 32
24的电子组态为: L,m,1,0,1S3S3P,l16l
11 S,m,,,1,s22s
(倒转次序) J,L,S,2
3 故基态为: P2
1225L,1,S,P 的电子组态为: ,倒转次序,故基态为 Cl3S3P31722
第五章
3512.4,105-1 V,,10(V)0.124
162,,0.248,10(z,1)5-2 利用 k,
22c?(,1),,17.66,10Z16 ,,0.248,10k,
解得 (Z,1),42?Z,43
3312.4,105-3 (应是电离一个L电子所做的功) E,,6.53,10(eV)L1.9
31?KK5-4 的末态,而的末态 J,?2J,1,4J,,2J,1,2,1,222
KK谱线强度同末态数成正比,故比强2倍。 ,1,2
3hC12.4,10E,,,,,,87.9(KeV)K1) 5-5 (,0.141k,
3hC12.4,10,(0.167,0.141)E,,,,,,13.6(KeV) L,0.167,0.141L,
,,hc(,)hC,KK,E,,,,,,3.01(KeV)M ,,,,M,KK,,
,,hc(,)hC,KK,E,,,,,,0.62(KeV)N ,,,,N,K,K,
3hc12.4,10L,,,,1.17(2)的波长(Å) L,3,EE,10.6,10ML
激发L系所需的最小能量为 ,E,E,E,13.6(KeV),L
2dsin,,n,5-6 利用
,5.4(Å) ?d,,,5.4,2sin2,0.5
2mC,0,5-7 利用散射光子能量最小。 ,,,,180h11,cos,,r
21mC0, ?h,,,,0.511,0.17(MeV)0.5111,1,0.5113
22,电子的动能 ThhmC ,,,,,e03
1154,,224242MeV PEmCmCmC?,,,(),1,,0.68()e000,,CCC33,,
,,,h,h,10,
,2,mC205-8 , 联立可得 2h,,20h,,5110,0,,h,,2mC,02,,h,,
10,100,4,2550 ?h,,,55.7(KeV)2
,,,hh,,5.7,
2,25.7mCmC,p2p5-9 , 可得 h,,5.7h,,,0,,h,,22mC,p2,,h,,
解得: h,,54.6(MeV)
5-10
222mCmC2mC2000,,h,,,,2mC222 02mCmCmC00011,cos,,,1,2,h,hh,
2, 故无论多大,不能产生正负电子偶。 h,,2mCh,0
5-11 采用质心系,反应后动量都为0,故只要考虑能量守恒。
22,h,,mC,mC(反应后电子在质心系动量为0。) ?,,00
故有,不可能产生光电效应。 m,m0
5-12 能量守恒 h,,E,E,,
,h222动量守恒 (),P,P,2PPcos,,,,,C
222(E,E),P,P,,,, ,?cos,2PP,,
2EEmC2,2,,0 ,,1224224EmCEmC2,,,00
故不能在真空中发生光子电子对过程。 ,
118,1)先考虑: 5-13 ((4d)S,m,,,1,s22ms
, L,m,2,1,3,l
s
13, 再同耦合: 5SS,m,1,,,S22
L,m,3,0,3,l
39j,3,, 22
48F满壳层缺2个,倒转次序,故基态为 ?4d92
32,E,RhC,(45,0.9),19.8(KeV)(2) k,4
h,19.8 ?,,,0.038,2511mC0
2,(h),,,h(1,cos) 故E,, e2,,1,(1,cos)2mC,h,0
119.8,0.038,22 ,,3.76,10(eV)11,0.038,2
,,,x52.5,,,,x,'x?x,x,,0.3,20.6(cm)I,Ie(3), ()0x0.765,'
,,',32.32ln10,(,),,x?x,,,8.3,105-14 (克/厘米) 325,48,,
第六章
12.26U,10eV6-1 , 当时,(Å) ,,,3.88V
U,100eV 时,(Å) ,,1.23
U,1000eV时,(Å) ,,0.388
,per6-2 (1) ,,1p,er
22hhCPe,(2) E,,,,Ehre22me,2m,ree
28mCE8,0.511er ?,,,340EhC0.012e
22216-3 (1) mC,mC,mC?,2v,0.866C002,1,
3hCh1.24,10,,,,,0.014(2)(Å) 6ppC0.833,10
1,,2dsin,,2,1.8,,1.86-4 (Å) 2
222p(hC)h?E,,,,0.025(eV)222 2m,,2m()2mCp
11222242226-5 由 得 E,cp,mCp,E,E,E(E,2mC)kk000CC
hhC ?,,,2pE(E,2mC)kk0
hC, eV12.262meV1,,02V2mCr0
eV,6 其中 V,V(1,),V(1,0.978,10V)r22mC0
224E,mC,pE02c,,,(),16-6 (1) 2,mCEmC000
E2(2) 故 E,(2,1)E,0.212(MeV)(),1,1?E,2EK00E0
,,1,9,,6-7 ~,?,t,,1.59,10(s),,,,,,4,t4,tC,4,C,
2,,,8{,t,,E,,,,t~,2,10(S)} ,2,C
,C197E~C,p~Cp~~,9.85(MeV)6-8 2,x2,10
x|y||z|,,,,,,,,,22abc|,|d,,Nedxedyedz6-9 (1) ,,,,,,,,,,
yabcxz,,,2abc,8Nedxedyedz ,,,000
218Nabc1N,,?, 8abc
axx1111,,aaaedxae,,(),(1,) (2) 0,0aae222
yz||||z,,bcbcy111,,,,2bcbcedyedzedyedz (3) ,,(1,),,,,00,,bcbcbce4
26-10 (1) |,(1,0,0;0,1,1)|
2(2) |,(x,y,z;0,0,0)|dxdydz111111,,,
1111112(3) |,(x,y,z;x,y,z)|dxdydzdxdydz111222111222,,,,,,000000
2n,6-11 设势阱边界为[0,a],则 (x),sinx,naaaaa2121n,n,a2(sin)(1cos),,,,,xxxdxxxdxxdx ,,,0002aaaaa
222 ?(x,x),x,x
aa2n12n,,222而 x,xsinxdx,x(1,cosx)dx,,00aaaa3axn112,2a ,,xxdx()cos0,0aaa3
2aaanan12,12,2a ,,,xx,,xxdx[sin]2sin0,0anaana322,,
2aaan22, ,,xxdxsin,0ana32,
222a,,a1a2n1a2naaa [xcosx]cosxdx ,,,,,,,022,03,n2,nan,2n,a32n,
222226aaaaa2()(1)?,,,,,,,,,,xx n,,22223412122,,nn
1按经典理论,,粒子在空间出现的几率相同为 p(x), ?U,0a
aa1a()故 ,,,xpxxdxxdx,,002a
2aa1a222(),,,xxpxdxxdx ,,003a
222aaa22 与时的量子结论一致。 n,,x,x,,,3412
r3,a211R(r),2()e6-12 10a1
r3r,a2211R(r),2()(1,)e 212a12a1
22d|rR|10由 及二阶导数小于0得 r,a(第一玻尔半径) ,0m1dr
22d|rR|21由 及二阶导数小于0得 r,4a,0m1dr
222,222r114eea,2*20a0ee4ederdr,,,,,,,,,,,,,,,,,6-13 33,,,,ra004rara,00
6-14 (略)
2,d(x)2mE2,,(x),06-15 立方程: ?: 222,dx
2,d(x)2m(V,E)30,,(x),0?: 322,dx
方程解为: ?: ,(x),01
2mE2,(x),AsinKx,BcosKxK,?: 22222,
mVE2(,)2,Kx03xCeK,(),,?: 332,利用波函数
标准
excel标准偏差excel标准偏差函数exl标准差函数国标检验抽样标准表免费下载红头文件格式标准下载
条件: 得B=0 x,0,(x),,(x)12
,Ka3 ---------(1) x,a,(x),,(x)AsinKa,Ce232
,Ka3,, ---------(2) ,(x),,(x)AKsinKa,,CKe23223
V,EK(1)10,2得: 或 tgKa,,Ka,n,,sin22(2)K3V0
V,E1,0由图解法, 可判断存在束缚态的条件为: y,Ka,y,n,,sin122V0
2222mVh,,,02 即 Va,,,028m32m,
2,2m(V,E)D0,p~e6-16 由透射几率 估计,由于分子都是宏观量,,
?p~0
第七章
27-1 根据原子的质量求 B,,mC
402 Ca:,m,40.3298,39.96259,0.36721,B,,mC,0.36721,931,342(MeV)
342B B,,,8.55(MeV)A40
562 Fe:,m,56.4634,55.9349,0.5285,B,,mC,0.5285,931,492(MeV)
492B B,,,8.79(MeV)A56
,33,1,6740A,,107-2 ,12.3,10(S),0.33,10(Ci)60
3,18,112.3,10A,,,,4.87,10(S)23 1N,6.02,10238
,18,79ln2故(年) T,,0.693/4.87,10/3.15,10,4.5,10,
,127-3 1克碳中碳14的含量为(克) M,1.3,10
236.02,10,1210故(个)/克 N,,1.3,10,5.59,10014
0.69310,1,1,1,1A,,5.59,10,0.21(sg),13(mg)故 075730,3.15,10
,1,1300A,,3(mg)而 100
ttA,,,tTT130AAeA由 ,,2?2,,,4.3300A3
ln4.331.47(年) ?t,()T,,T,2.12,5730,12147ln20.693
1,7-4 由天,/天 ?,,0.1,,10,
,,t N,Ne0
54,NN(4),N(5),,1010,,e,e,0.064 NN00
,37-5 ,M,226.0254,222.0176,4.0026,5.2,10(u)
2,3 ,MC,5.2,10,931,4.84(MeV)
222E,,4.84,4.75(MeV)故 ,226
4E,5.30,(1,),5.403(MeV)7-6 00206
4 E,4.50,(1,),4.587(MeV)01206
故E,E,E,0.816(MeV)r0001
,7-7 变化时放出的能量为 1.89,1.02,2.91(MeV),
2K电子的结合能 ,~RhC(23,1),0.007(MeV)k
故中微子的能量为 E,2.91,0.007,2.903(MeV),
,37-8 (1) ,M,4.002603,14.00307,16.99913,1.007825,,1.282,10(u)
2,3? 反应能 Q,,MC,,1.282,10,931,,1.194(MeV)
,3(2) ,M,1.007825,9.012183,6.015123,4.002603,2.28,10(u)
2,3? 反应能 Q,,MC,2.28,10,931,2.13(MeV)
,37-9 ,M,1.007825,3.01605,1.008665,3.016029,,8.19,10(u)
2,3 ?Q,,MC,,8.19,10,931,,0.763(MeV)
3,1入射质子的能量 E,,0.763,1.017(MeV)p3
1132又 Q,(1,)E,(1,)E?E,[Q,E],0.928(MeV)npnp3343
而 E,Q,E,E,,0.763,0.928,3,1.309(MeV)Henp
7-10 参阅史包尔斯基采用原子质量的精确表达式:
1A22233M,1.007825Z,1.008665(A,Z),aA,aA,a(,Z)/A,0.000627Z/A 1232
1A,M3,0.00084,2a(,Z)/A,0.000627,2,Z/A,0 由得, ,032,Z
2A(0.00084,a)33 ?Z,,0.001254A2a3
由相当衰变最稳定的那些原子的Z、A代入上式可求得 ,a,0.0833
AZ, 故 232,0.015A
7-11 (1)设碳原子的质量为,碰撞后的速度为,能量为,由 MvEAAA
1122,,E,E,E,mv,Mv 能量守恒 AnnAA0022
,,P,P,P,mv,mv 动量守恒 0AnnAA
2mn,vvA0 可得 ,mMnA
2m11n222?,,EMvM()v AAAA0,22mMnA
MA41m2n,mv,{}n0 m,M2nA2()mn
2A,(1),E, (A为碳原子的质量数) [1]02A,(1)
A,12因此经一次碰撞后的中子能量为 ()E,0.72E00A,1
N故N次碰撞后的能量为 0.72E0
NN(2) 0.72E,0.72,2,0.025(eV)0
log0.01251.9?N,,,13.6~14 (次) log0.720.14
7-12 (1)质子的结合能
B,18.99840,1.007825,19.99244,0.013785,931.5,12.8407(MeV)p
1920c 的激发能级 E,B,E,12.8407,224.4,,13.0539(MeV)Ne1pp20
19c E,B,E,12.8407,340.4,,13.1641(MeV)2pp20
19c E,B,E,12.8407,873.5,,13.6705(MeV)3pp20
19c E,B,E,12.8407,935.3,,13.7292(MeV)4pp20
19c E,B,E,12.8407,1085.0,,13.8715(MeV)5pp20
,,C197,20,t,,,,10(s)(2)由 可得各能级的平均寿命为: 2,E2C,E6,E
197,20,19 ,t,,10,3.28,10(s)16,1.0
197,20,20 ,t,,10,7.29,10(s)26,4.5
197,20,20 ,t,,10,6.31,10(s)36,5.2
197,20,20 ,t,,10,4.10,10(s)46,8
197,20,20 ,t,,10,8.21,10(s)56,4
D,T,,,n,17.6MeV7-13
31023261千克氚释放的能量为 17.6,,6.02,10,35.3,10(MeV)3
97100万千瓦电站一年中释放的能量为 焦耳/秒×3.15×10秒 10
217 ,6.25,10,3.15,10(MeV)
29,1.97,10(MeV)
29101.97,,55.8故一年中消耗的氚为 (千克) 2635.3,10
975810,3.15,10,9.55,10折合煤为 (千克)(吨) ,9.55,1073.3,10
7.86n,2423r7-14 P,,6Nt,2.5,10,0.1,,6.02,10,0.021n56
34,34-111,7-15 (月)即要有个质子 ,,,6.94,106.94,1032,1.2,10,12
34116.94,10才能每月观察到一次衰变,它相当于(克) ,1.15,10236.02,10
1811115所需的水为 (克)(吨) x,1.15,10,,10.35,10,10.35,102