Maxwell Distribution: Deriving Mean and Max Speed

[blueyellow]

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)

Homework Equations

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
please help, thanks in advance

 

[Redbelly98]

Do you have an expression for the distribution function? How would normally find the maximum of a function?

 

[blueyellow]

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]

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.

 

[blueyellow]

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

 

[Redbelly98]

The mean of a quantity is often calculated by using it's probability function and setting up an integral ... are you familiar with that method?