5.281. Let us write the energy and momentum conservation laws
in the reference frame fixed to the electron for the moment preceding
the collision with the photon: hco moc2 &plc = my, where
m = mo 1/ 1 — (v/c)2. From this it follows that v = 0 or v = c.
The results have no physical meaning.
5.282. (a) Light is scattered by the free electrons; (b) the increase
of the number of electrons that turn free (the free electrons have the
binding energy much lower than the energy transferred to them by
the photons); (c) the presence of a non-displaced component is due
to scattering by the strongly bound electrons and the nuclei.
5.283. a = 42dc [sin (02/2) — t1 sin (01/2)1/(1 — 1) = 1.2 pm.
5.284. T = hcoril(1
i) = 0.20 MeV.
5.285. (a) e.)' = 2nc/(k 2nh/mc) = 2.2.1020 rad/s;
2nch/X
(b) T — 1+ Xmc 12rch
60 keV.
hco
5.286. ho)' = 1+2 (h(o/mc2) sin (0/2) 0.144 MeV.
5.287. sin (0/2) =1/ mc (p — p')I2pp' . Hence 0=120°.
5.288. hu) = [1 +V 1 + 2mc2IT sin2 (0/2)] T/2 = 0.68 MeV.
5.289. = (2mh/mc) (1/1 + 2mez/T,,,ox — 1) = 3.7 pm.
5.290. tan
1/4h/mcA— 1
cp
n
A.
1+hoon,2
,
31°.
5.291. p
= 21 (1+ ri)
(1 ± 211) 03
=. 3.4 cm.
5.292. A?. ---- (4hImc) sin2 (0/2) = 1.2 pm.
6.1. r = 3e2/2E = 0.16 nm, 2 = (2acle) y mr3 = 0.24
6.2. b = 0.73 pm.
6.3. (a) rmin = 0.59 pm; (b) r m in = (2Ze2I T) (1 -1- mc,ImLi)
0.034 pm.
6.4. (a) pmin = (Ze 2/T) cot2 (0/2) = 0.23 pm;
(b)
rmin =- _
[1
cosec (0/2)1 Ze2IT = 0.56 pm.
6.5. p;---e, 2 1/2mT/[1+(2bT/Ze2 )21.
6.6. T e =mpe4Imeb2T = 4 eV.
n sin (0/2)
6.7. b —
where n =1/1 +Uo/T.
- 1/ 1+ n2 — 2n cos (0/2)
6.8. (a) cos (0/2) = b/(R
r); (b) dP = 1/2 sin 0 d0; (c) P
= 1/2.
6.9. 3.3-10-5.
6.10. d = (4Jr2T2/nIZ2e4) sin4 (0/2) = 1.5 p,m, where n is the
concentration of nuclei.
6.11. Zpt = ZAg T1APt/AAg = 78.
6.12. (a) 1.6.106; (b) N = and (Ze2I T)2 cot2 (0012) IOv = 2.0.107,
where n is the concentration of nuclei.
6.13. P = and (Ze2/mv2)2 = 0.006, where n is the concentration
of nuclei.
6.14. ANIN = 1 — nnZ2e4/T2 tan2 (00 /2) = 0.6.