Question

By differentiating the expression for the Maxwell-Boltzmann energy distribution, show that the peak of the distribution occurs at an energy of (1/2)kT.

Answer 1

We know that the Maxwell- boltzmann energy distribution function is given by
$f (v)= C_{1}v^{2}*e^{-C_{2}v^{2}}$
Where
C₂ = (m/2kT)
Now on differentiating the above equation
$f '(v)= \left [ C_{1}v^{2}*e^{-C_{2}v^{2}}*\left ( -2C_{2}v \right ) \right ]+\left [ e^{-C_{2}v^{2}}*2C_{1}v \right ]$
$f '(v)= \left [ -2C_{1}C_{2}v^{3}*e^{-C_{2}v^{2}} \right ]+\left [ 2C_{1}v*e^{-C_{2}v^{2}} \right ]$
For the maximum value put
$f '(v)= 0]$
$0= \left [ -2C_{1}C_{2}v^{3}*e^{-C_{2}v^{2}} \right ]+\left [ 2C_{1}v*e^{-C_{2}v^{2}} \right ]$
$\left [ 2C_{1}C_{2}v^{3}*e^{-C_{2}v^{2}} \right ]=\left [ 2C_{1}v*e^{-C_{2}v^{2}} \right ]$
$v^{2} = \frac{1}{C_{2}}$
Now putting the value of C₂
$v^{2} = \frac{2kT}{m}$
$v= \sqrt{\frac{2kT}{m}}$
So the peak of the curve will be at this velocity.