1000 Solved Problems in Modern Physics

4.3 Solutions 259

of mass these two collisions appear to be equivalent so thatc′=c. We can

then write

f(ν 1 )f(ν 2 )=f(ν 3 )f(ν 4 )

or lnf(ν 1 )+lnf(ν 2 )=lnf(ν 3 )+lnf(ν 4 )(1)

Since kinetic energy is conserved

ν 12 +ν^22 =ν 32 +ν^24 (2)

Equations (1) and (2) are satisfied if

lnf(ν)∝ν^2 (3)

orf(ν)=Aexp(−αν^2 )(4)

whereAandαare constants. The negative sign is essential to ensure that no

molecule can have infinite energy.

LetN(ν)dνbe the number of molecules per unit volume with speedsνto

+dν, irrespective of direction. As the velocity distribution is assumed to be

spherically symmetrical,N(ν)dνis equal to the number of velocity vectors

whose tips end up in the volume of the shell defined by the radiiνand+dν,

so that

N(ν)dν= 4 πν^2 f(ν)dν (5)

Using (4) in (5)

N(ν)dν= 4 πAν^2 exp(−αν^2 )(6)

We can now determineAandα.IfNis the total number of molecules per

unit volume,

N=

∫∞

0

N(ν)dν (7)

Using (6) in (7)

N= 4 πA

∫∞

0

ν^2 exp(−αν^2 )dν= 4 πA(1/4)(π/α^3 )^1 /^2

orN=A(π/α)^3 /^2 (8)

IfEis the total kinetic energy of the molecules per unit volume

E=

1

2

m

∫∞

0

ν^2 N(ν)dν=

4 πAm

2

∫∞

0

ν^4 exp(−αν^2 )dν

orE=(3mA/4)(π^3 /α^5 )^1 /^2 (9)

where gamma functions have been used for the evaluation of the two integrals.

Further,