Irodov – Problems in General Physics

5.251. t = (i 3 — 1) cpd/18crT: = 3 hours, where c is the spe-
cific heat capacity of copper, p is its density.

5 252. T2 == T1 d121= 0.4 kK.

5.253. (a) Cv = (OU/OT)v =16 aT^3 VIc=3 nJ/K, where U=

.4aTIVIc; (b) S = 16crT 3 V/3c = 1.0 nJ/K.

5.254. (a) cop,. = 3Tla = 7.85.10" s- 1 ; (b). X•pr = 2nca/5T =

= 1.44 ti,m.

5.255. (a) u. = (kT 17E 2 0)(0 2 ; (b) u. = (h/n 2 c 3 ) (0^3 e-5(0^47 '.

5.256. — ieuclh

vs

— 2 /kTi,i

167C2Cnk-3

c 3 e^22 - thvillT —^1 afic •

5.257. AP = 4n2c2hT50%/b6 (e2nitc/kb• 1) = 0.31 W/cm^2 , where

b is the constant in Wien's displacement law.

5.258. (a) 1.1 [tm; (b) 0.37; (c) P^2 /P 1 =--- (T 2 /T 1 ) 4 (1 — Y2)/(1—Y1) =

= 4.9.

5.259. rc. dm —

co 2 dw

dX

8a-4

T

1

oo elm' hr • -^ e2nhoira,

5.260. (a) (j) = Pk181c^2 chr^2 = 6.10^13 cm-2s-1;

(b) r = VP2t./2An/2ric = 9 m.

5.261. dpIdt = (De/c.

5.262. (p) = 4 (1 -I- p) Elnd^2 c-r^ 50 atm.

5.263. p = (Etc) p^2 -I- 2p cos 20 = 35 nN • s.

5.264. p = (17c) (1 + p) cost 0 = 0.6 nN/cm 2.

5.265. F = nR 2 Ilc = 0.18 IAN.

5.266. F = P/2c (1 +^1 2).

5.267.( a) Ap p (b) -

2hw • I 1 —02 A

P=

2hw Here

1-0 • R=

= V/c. It is evident that in the reference frame fixed to the

mirror the latter obtains the smaller momentum.

5.268. sin (8/2) Elmc F, 0 = 0.5°.

5.269. Doi/coo = — (1 — e-vm/Rc^2 ) < 0, i.e. the frequency of

the photon decreases.

5.270. V = 23 - thc (1 — 1/11)/eAk = 16 kV.

5.271. V = nhcled sin cc = 31 kV.

5.272. ?min = 2Tih/mc (y —1) = 2.8 pm, where v =1 /17.^1 — (v/c)^2.

5.273. 332 nm, 6.6.10 5 m/s.

5.274. A— 2Ttch(112-2, '21)'1) =1.9 eV.^

5.275. Tmax 4.4 ( 1 \ 7  I )

5.276. Tmax = h ((Do + (0) — Af = 0.38 eV.

5.277. w = 2nchTleA, = 0.020.

5.278. v„. 4 „x = 6.4.10 5 m/s.

5.279. 0.5 V; the polarity of the contact potential difference

is opposite to that of external voltage.

5.280. hlmc, the Compton wavelength for the given particle.

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