Irodov – Problems in General Physics

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6.186. 0 = (h/k) 13 /183-t 2 no/(0- 13 + 2v-2) = 470 K, where no is the
concentration of the atoms.


.4 0 0 c^0 0 w H^ C^ C^ H
wz
(0,1

EM 01--off-lb (0 2

(a)

wk
cos

(b)
Fig. 46.

6.187. v kWh r6n 2 no = 3.4 km/s, where no is the concent-
ration of the atoms. The tabulated values are: vii = 6.3 km/s,
v 1 =3.1 km/s.
6.188. The oscillation energy of a mole of a "crystal" is
8/T
T 12 x dx
k0

(^3)
ex-1P
0
where x---hcolkT. Hence the molar heat capacity is
9/T
C R (2T x dx^ 81T
8 J ex —^1 ee/T __ 1 •
0
When T >> 0, the heat capacity C R.
6.189. (a) dN/do 21Ina 1 1 — co 2 ; (b) N = 11 a, i.e.
is equal to the number of the atoms in
dll (^) the chain.
dm (^)
6.190. U, = 9R0/8μ = 48.6 J/g,
8/2 whereμ is the molar mass of copper.
10- 6.191.^ (a) 0^ 220 K; (b) C
10 J/(mol•K); (c) comax = 4.1 x
X 1013 rad/s.
6.193. Yes, because the heat capac-
5 - ity is proportional to T 3 at these tern-
epi peratures.
6.194. (E) = 3 /8 k0.
6.195. See Fig. 47.
0.5 (^10) 6.196. komax = 28 meV,
hkmax^'
g-cm/s.
Fig. 47. (^) 6.197. (^) (a)
T max = (3/i^2 n)^2 /^3 h^2 /2m;
(b) (7') = 3 /5Tmax-
6.198. ri = 1 — 2-3P X 65%.
6.199. 0.93.
6.200. z 3.10 4 K.
360

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