Question

The speed of a molecule in a uniform gas at equilibrium is a random variable whose probability density function is given by

$f\left(x\right)=\left\{\begin{array}{ll}a{x}^2{e}^{-b{x}^2}& x\ge 0\\ 0& x<0\end{array}\right.$

where$b=m/2kT$and$m$denote, respectively,

Boltzmann’s constant, the absolute temperature of the gas,

and the mass of the molecule. Evaluate $a$in terms of$b$.

Short answer

Therefore $a$in terms of$b$then$a=\frac{4b{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\sqrt{\mathrm{\pi }}}$.

Step-by-step solution

Given Information.

$f\left(x\right)=\left\{\begin{array}{ll}a{x}^2{e}^{-b{x}^2}& x\ge 0\\ 0& x<0\end{array}\right.$

Where,

$b=m/2kT$

Explanation.

$$\begin{gathered} {\int }_0^{\infty }f\left(x\right)dx=1 \\ {\int }_0^{\infty }a{x}^2{e}^{-b{x}^2}dx=1 \end{gathered}$$

localid="1649615141474" $$\begin{gathered} b{x}^2=t \\ 2xdx=dt \\ {\int }_0^{\infty }a{x}^2{e}^{-t}\times \frac{dt}{2bx}=1 \end{gathered}$$

Explanation.

$$\begin{gathered} b{x}^2=t\Rightarrow x^2=\frac{t}{b} \\ \Rightarrow x=\sqrt{\frac{t}{b}} \\ \frac{a}{2b^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{\int }_0^{\infty }{t}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{e}^{-t}dt=1 \\ \frac{a}{2b{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\times \sqrt{\frac{3}{2}}=1 \\ \frac{a\sqrt{\mathrm{\pi }}}{4b{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=1 \\ a=\frac{4b{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{\sqrt{\mathrm{\pi }}} \end{gathered}$$