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**Unformatted text preview: ** 0 18.80: a) m f (v ) dv = 4 2kT m = 4 2kT 3 2 v
0 2 e -mv 2 / 2 kT dv 32 1 4(m 2 KT m 2 KT = 1 where the tabulated integral (given in Problem18.81) has been used. 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) dv must be 1. ...

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