Chapter 1, Problem 1B.3P

Interpretation Introduction

Interpretation: The mean speed of the emerging beam relative to the initial value has to be calculated.

Concept introduction: The expression for the mean speed is derived from Maxwell Boltzmann distribution.  The mean speed for the gas particles is represented as,

  $v_{\text{mean}}={\left(\frac{8RT}{\pi M}\right)}^{1/2}$

Answer to Problem 1B.3P

The speed of the emerging beam relative to the initial value is $\underset{\_}{v_x=0.47{\langle v_x\rangle }_{initial}}$.

Explanation of Solution

The expression for the velocity of the emerging beam is,

  $v_{\text{x}}=K{\displaystyle \underset{0}{\overset{a}{\int }}{v}_{\text{x}}f\left(v_{\text{x}}\right)d{v}_{\text{x}}}$                                                                                       (1)

Where,

• • $K$ is the constant of proportionality.
• • $v_{\text{x}}$ is the velocity.

The value of $f\left(v_{\text{x}}\right)$ is given by,

  $f\left(v_{\text{x}}\right)={\left(\frac{m}{2\pi kT}\right)}^{1/2}{e}^{-m{v}^2/2kT}$

Where,

• • $m$ is the mass.
• • $k$ is the Boltzmann constant.
• • $T$ is the temperature.

The normalised function is written as,

  $\begin{array}{c}\frac{v_{\text{x}}}{{\displaystyle \underset{0}{\overset{a}{\int }}{v}_{\text{x}}d{v}_{\text{x}}}}=K{\displaystyle \underset{0}{\overset{a}{\int }}f\left(v_{\text{x}}\right)}d{v}_{\text{x}}\\ 1=\text{}K{\displaystyle \underset{0}{\overset{a}{\int }}f\left(v_{\text{x}}\right)}d{v}_{\text{x}}\end{array}$

Substitute the value of $f\left(v_{\text{x}}\right)$ in the above equation.

    $1=K{\left(\frac{m}{2\pi kT}\right)}^{1/2}{\displaystyle \underset{0}{\overset{a}{\int }}{e}^{-m{v}^2/2kT}}d{v}_{\text{x}}$                                                                       (2)

The equation (2) is evaluated by defining an error function.

  $\begin{array}{c}\frac{m{v}_{\text{x}}^2}{2kT}=n^2\\ v_{\text{x}}^2=\frac{2kT}{m}{n}^2\\ v_{\text{x}}={\left(\frac{2kT}{m}\right)}^{1/2}n\end{array}$

Or,

  $d{v}_{\text{x}}={\left(\frac{2kT}{m}\right)}^{1/2}dn$                                                                                          (3)

Substitute equation (3) in equation (2) to find $n$.

  $1=K{\left(\frac{m}{2\pi kT}\right)}^{1/2}{\left(\frac{2kT}{m}\right)}^{1/2}{\displaystyle \underset{0}{\overset{b}{\int }}{e}^{-n^2}dn}$

Here, $b={\left(\frac{m}{2kT}\right)}^{1/2}\times a$

  $\begin{array}{c}1=K{\left(\frac{2kT}{m}\right)}^{1/2}{\left(\frac{m}{2\pi kT}\right)}^{1/2}{\displaystyle \underset{0}{\overset{b}{\int }}{e}^{-n^2}dn}\\ ={\left(\frac{K}{\pi }\right)}^{1/2}{\displaystyle \underset{0}{\overset{b}{\int }}{e}^{-n^2}dn}\end{array}$

According to error function,

  $\text{erf}\left(z\right)=\frac{2}{{\pi }^{1/2}}{\displaystyle \underset{0}{\overset{z}{\int }}{e}^{-n^2}dn}$

Therefore,

  $\begin{array}{c}1={\left(\frac{K}{\pi }\right)}^{1/2}{\displaystyle \underset{0}{\overset{b}{\int }}{e}^{-n^2}dn}\\ =\frac{1}{2}\sqrt{K}\text{erf}\left(b\right)\\ =\frac{2}{\text{erf}\left(b\right)}\end{array}$

To calculate the mean velocity of the beam, substitute the values in equation (1).

    $\begin{array}{c}{v}_{\text{x}}=K{\left(\frac{m}{2\pi kT}\right)}^{1/2}{\displaystyle \underset{0}{\overset{a}{\int }}{v}_{\text{x}}{e}^{-m{v}_{\text{x}}^2/2kT}d{v}_{\text{x}}}\\ =K{\left(\frac{m}{2\pi kT}\right)}^{1/2}\left(\frac{-kT}{m}\right){\displaystyle \underset{0}{\overset{a}{\int }}\frac{d}{d{v}_{\text{x}}}\left(e^{-m{v}_{\text{x}}^2/2kT}d{v}_{\text{x}}\right)}\end{array}$

    $v_{\text{x}}=-{\left(\frac{kT}{2m\pi }\right)}^{1/2}\left(e^{-m{v}_{\text{x}}^2/2kT}-1\right)$ (4)

The initial velocity, $a={\langle v_{\text{x}}\rangle }_{\text{initial}}$ and the initial velocity is ${\left(2kT/m\pi \right)}^{1/2}$.  The expression for average magnitude of one dimensional velocity in the $x$ direction is,

    $\begin{array}{c}{\langle v_{\text{x}}\rangle }_{\text{initial}}=2{\displaystyle \underset{0}{\overset{\infty }{\int }}{v}_{\text{x}}f\left(v_{\text{x}}\right)}d{v}_{\text{x}}\\ =2{\displaystyle \underset{0}{\overset{\infty }{\int }}{v}_{\text{x}}{\left(\frac{m}{2\pi kT}\right)}^{1/2}}{e}^{-m{v}^2/2kT}d{v}_{\text{x}}\\ ={\left(\frac{m}{2\pi kT}\right)}^{1/2}\left(\frac{2kT}{m}\right)\\ ={\left(\frac{2kT}{m\pi }\right)}^{1/2}\end{array}$

For $a=\infty$ the value of constant, $b=\infty$ and the value of error function, $b\text{erf}\left(b\right)=1$. Substitute the value of $a$ and $b\text{erf}\left(b\right)=\text{erf}\left(1/{\pi }^{1/2}\right)$ in equation (4).

  $v_{\text{x}}={\left(\frac{2kT}{m\pi }\right)}^{1/2}\times \frac{1-e^{-1/\pi }}{\text{erf}\left(1/{\pi }^{1/2}\right)}$                                                                           (5)

The value of $\text{erf}\left(1/{\pi }^{1/2}\right)$ is,

  $\begin{array}{c}\text{erf}\left(1/{\pi }^{1/2}\right)=\text{erf}\left(\text{0}\text{.56}\right)\\ =0.57\end{array}$

The value of $e^{-1/\pi }$ is $0.73$.

Substitute the values in equation (5) to calculate mean velocity.

    $\begin{array}{c}{v}_{\text{x}}={\left(\frac{2kT}{m\pi }\right)}^{1/2}\left(\frac{1-0.73}{0.57}\right)\\ =\underset{\_}{0.47{\langle v_x\rangle }_{initial}}\end{array}$

Thus, the value of mean velocity of emerging beam is $\underset{\_}{v_x=0.47{\langle v_x\rangle }_{initial}}$.