# Maxwell distribution

## Homework Statement

show that the mean speed of the particles (v-bar) and the speed of the maximum of the distribution are given by

v-bar=((8kT)/(pi*m))^(1/2)
and
v-max=((2kT)/m)^(1/2)

## The Attempt at a Solution

tried looking thru noted and textbook. i sussed that one is supposed to use standard integrals, but no notes or textbook is telling me how or where

Redbelly98
Staff Emeritus
Homework Helper
Do you have an expression for the distribution function? How would normally find the maximum of a function?

thanks for the suggestion
ok, so the distribution function is
p(v) is proportional to v^2 exp(-mv^2 /2kT)
so i differentiate it and get
dp/dv is proportional to (-m/2kT)*2v^3 exp(-mv^2 /2kT)=0
hmm

Redbelly98
Staff Emeritus
Homework Helper
Not a bad start, that's the right idea, but you didn't differentiate correctly.

p(v) is proportional to v2·e-mv2/2kT,
or in other words (a function of v)·(another function of v).

Use the product rule for differentiation to find the derivative.

i got it
thanks a lot
still clueless on the mean speed tho

Redbelly98
Staff Emeritus