Question

Maxwell Boltzmann Distribution [Most Probable Velocity]

Answer 1

I'm supposed to take the derivative of this to find the most probable velocity but I'm not sure where i'm messing up on the math

Answer 2

\[f'(v) = 4\pi(\frac{ m }{ 2*\pi kT })^{3/2} * [v^{2}e^{-mv/kT}+2ve^{\frac{ -mv^{2} }{ kT }}]\]

Answer 3

product rule!

Answer 4

**Product** rule:

\(f'(v) = 8 \pi \left( \dfrac{m}{2 \pi kT} \right)^{3/2} v \cdot e^{- mv^2 / 2kT} + 4 \pi \left( \dfrac{m}{2 \pi kT} \right)^{3/2} v^2 \cdot \dfrac{- 2mv}{2kT} e^{- mv^2 / 2kT} \)

set that to zero and cancel out a whole load of stuff and it simplifies somewhat miraculously to the answer

Answer 5

something like this? ;;

\[[4\pi( \frac{ m }{ 2\pi*kT })^{3/2} *ve^\frac{ -mv^2 }{ 2kT }][2+v(\frac{ -mv }{ kT })]\]

Answer 6

yep, that gives the solution !! as in the term on the right = 0 \(\implies\) the accepted answer.

i just hacked away at terms without thinking. to be über strict, you might look for other null solutions, is can you just cancel all that stuff in the left hand term? and is it a min or a max ?!?!

i shoulda said this right at the start to set it up for product rule.....

\(f(v) = \color{red}{ 4 \pi \left( \dfrac{m}{2 \pi kT} \right)^{3/2} v^2} \cdot \color{blue}{ e^{- mv^2 / 2kT}}\)

but that's what you've done....

Answer 7

so I end up getting 2 + (-mv^2)/KT = 0

solving for v I get v = sqrt(2KT/m) which I believe is the answer?

Answer 8

that's what my notes say anyway ;;

Answer 9

\(v = \sqrt{\dfrac{2kT}{m}} \qquad \huge{\color{green}{\checkmark}}\)

Answer 10

thanks, I'll have to review this tonight