1000 Solved Problems in Modern Physics

4.3 Solutions 261

4.5 (a)νpis found by maximizing the Maxwellian distribution.

d

dν

[ν^2 exp(−mν^2 / 2 kT)]= 0

exp(−mν^2 /kT)[2ν−mν^3 /kT]= 0

whenceν=νp=(2kT/m)^1 /^2

(b)νp:<ν>:<ν^2 >^1 /^2 :: (2kT/m)^1 /^2 :(8kT/πm)^1 /^2 :(3kT/m)^1 /^2

=

√

2:

√

8 /π:

√

3

4.6 <ν^2 >^1 /^2 =

(

3 kT

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 × 105

0. 3

) 1 / 2

=632 m/s

4.8<

1

ν

>=

1

N

∫∞

0

1

ν

N(ν)dν

=

1

N

∫∞

0

1

ν

. 4 πN

( m

2 πkT

) 3 / 2

v^2 exp(−mν^2 / 2 kT)dν

Setmν^2 / 2 kT=x;vdν=kTdx/m

<

1

ν

>=(2m/πkT)^1 /^2

∫∞

0

exp(−x)dx=(2m/πkT)^1 /^2

4.9 N(ν)dν= 4 πN(m/ 2 πkT)^3 /^2 ν^2 exp(−mv^2 / 2 kT)dν (1)

νp=(2kT/m)^1 /^2 (2)

Letν/νp=α;dν=νpdα (3)

Use (2) and (3) in (1)

N(α)dα=

4 N

√

π

α^2 exp(−α^2 )dα

4.10 Fraction

f=

N(ν)dv

N

= 4 π

[ m

2 πkT

] 3 / 2

ν^2 exp(−mν^2 / 2 kT)dν

ν=

199 + 201

2

=200 m/s

dν= 201 − 199 =2m/s