Irodov – Problems in General Physics

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 moc 2 &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 (0 2 /2) — t1 sin (0^1 /2)1/(1 — 1) = 1.2 pm.
5.284. T = hcoril(1 i) = 0.20 MeV.
5.285. (a) e.)' = 2nc/(k 2nh/mc) = 2.2.10 20 rad/s;
2nch/X
(b) T — 1+ Xmc 12rch 60 keV.

hco

5.286. ho)' = 1+2 (h(o/mc 2 ) 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 + 2mc 2 IT sin 2 (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) sin^2 (0/2) = 1.2 pm.
6.1. r = 3e 2 /2E = 0.16 nm, 2 = (2acle) y mr 3 = 0.24
6.2. b = 0.73 pm.
6.3. (a) rmin = 0.59 pm; (b) rm i n = (2Ze^2 I T) (1 -1- mc,ImLi)
0.034 pm.
6.4. (a) pmin = (Ze 2 /T) cot 2 (0/2) = 0.23 pm; (b)^ rmin =-_
[1 cosec (0/2)1 Ze^2 IT = 0.56 pm.
6.5. p;---e, 2 1/2mT/[1+(2bT/Ze 2 ) 2 1.
6.6. T e=mpe 4 Imeb 2 T = 4 eV.
n sin (0/2)
6.7. b — where n =1/1 +Uo/T.

• 1/ 1+ n 2 — 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 = (4Jr 2 T 2 /nIZ 2 e 4 ) sin 4 (0/2) = 1.5 p,m, where n is the
  concentration of nuclei.

6.11. Zpt = ZAg T1APt/AAg = 78.
6.12. (a) 1.6.10 6 ; (b) N = and (Ze 2 I T) 2 cot 2 (0 0 12) IOv = 2.0.10^7 ,
where n is the concentration of nuclei.
6.13. P = and (Ze^2 /mv^2 )^2 = 0.006, where n is the concentration
of nuclei.
6.14. ANIN = 1 — nnZ^2 e^4 /T^2 tan^2 (0^0 /2) = 0.6.