Irodov – Problems in General Physics

(Joyce) #1

2.87. T — 2k (i mAvN m 0)2 ) — 0.37 kK.


2.88. v =

3kT 1n (m2/m1) =1.61 km/s.
M2 — 1
2.89. T = 113 mv 2 1k.
2.90. dN/N = (-72akT 2' ) 3/2 e-mv 212 hT 2nvl dvl dvx.

2.91. (vx)= 0, (I vx ) =1/.2kT I nm•
2.92. (v1) = kT/m.
2.93. v = 1 / 4 r1 (v), where (v) = li8kT/am.
Co
2.94. p = 2mvx • vx dn (vx) = nkT , where dn (vx) =

(m12nkT) 1/2n • e-m4/2hT dvx.
2.95. (1/v) =1/2mInkT = 4n (v).
2.96. dN/N= 21s (nkT)-3/^2 e-ena^ dc; ep, =^1 /^2 kT; no.
2.97. SNIN = 3 6n e-3 12 6i= 0.9%.
CO
2.90 — = Q AN 231
(nkT)^312 J

The principal contribution to the value of the integral is provided
by the smallest values of a, namely a x ao. The slowly varying factor
lii-can be taken from under the radical sign if ascribed the constant
value 1/ s- o. Then


AN 1 N = 2 V eolnkT e-sona
2.99. (a) vp,. = 1/3kT/m; (b) apr = kT.
00
2.100. dv = dn (42143-t)v cos 0 = n (2kT/am)I/2 sin 0 cos 0 de.
v=-0
n/2
2.101. dv = dn (d52/4n) v cos 0 = n (m/2akT)3/2 emv- 2 /^211 Tv^3 dv.
e=o
2.102. F = (kT I Ah) ln = 0.9.10-0 N.
2.103. NA = (6RT Ind 3 Apgh) ln 6.4.10 23 mo1-1.
2.104. 11/1 0 = ec.m2-m1vninT =^1.^39.^
2.105. h — kT (mIn (n21%) 2 — mi) g •
2.106. Will not change.
2.107. (U) = kT. Does not depend.
2.108. w riliT I M1 N 70 g.
2.109. M — 2RTp In 1
(P— Po) (r2—r?) 0)2 •
2.110. co = V (2RT M1 2 ) ln rl = 280 rad/s.
2.111. (a) dN = no e-ar 2 /hT4ar 2 dr; (b) kT/a; (c) dN =
= (a/akT) 3 / 2 e-ar2/14T4TEr2dr; (d) Will increase 13/2-fold.

1/^8 e-e/kT de.^
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