(a) Show that is the Maxwell–Boltzmann distribution of Eq. (18.32). (b) In terms of the physical definition of ƒ(?v?), explain why the integral in part (a) ?must ?have this value.

Solution 84P 3 mv2 a) (v)dv = 4( m )2 v e 2kTdv 0 2 kT 0 m 2 1 = 4( 2kT ) (4(2kT) 2kT = 1 b) f(v)dv is the probability that a particle has speed between v and v+dv, the probability that the particle has some speed is unity, so the sum (integral ) of f(v) must be 1