4.3 Solutions
261
4.5 (a) ν p is found by maximizing the Maxwellian distribution.
d
dν
[ν
2 exp(−mν
2
/2kT )] = 0
exp(−mν
2
/kT )[2ν − mν
3
/kT ] = 0
whence ν = ν p = (2kT /m)
1/2
(b) ν p :< ν >:< ν
2
>
1/2 :: (2kT /m)
1/2 : (8kT /π m)
1/2 : (3kT /m)
1/2
=
√
2 :
8/π :
√
3
4.6 < ν
2
>
1/2
=
3kT
m
1/2
=
3 × 1.38 × 10
−23
× 273
1.67 × 10 −27
1/2
= 2,601 m/s at N.T.P
< ν
2
>
1/2
=
3 × 1.38 × 10
−23
× 400
1.67 × 10 −27
1/2
= 3,149 m/s at 127
◦ C.
4.7 < ν
2
>
1/2
=
3 p
ρ
1/2
=
3 × (300/760) × 1.013 × 10
5
0.3
1/2
= 632 m/s
4.8 <
1
ν
>=
1
N
∞
0
1
ν
N (ν) dν
=
1
N
∞
0
1
ν
.4π N
m
2πkT
3/2
v
2 exp(−mν
2
/2kT ) dν
Set mν
2
/2kT = x; vdν = kT dx/m
<
1
ν
>= (2m/π kT )
1/2
∞
0
exp(−x) dx = (2m/π kT )
1/2
4.9 N (ν)dν = 4π N (m/2πkT )
3/2
ν
2 exp(−mv
2
/2kT )dν
(1)
ν p = (2kT /m)
1/2
(2)
Let ν/ν p = α; dν = ν p dα
(3)
Use (2) and (3) in (1)
N (α)dα =
4N
√
π
α
2 exp(−α
2 )dα
4.10 Fraction
f =
N (ν)dv
N
= 4π
m
2πkT
3/2 ν
2 exp(−mν
2
/2kT )dν
ν =
199 + 201
2
= 200 m/s
dν = 201 − 199 = 2 m/s