Chapter 39, Problem 51PQ

(a)

To determine

The rest mass energy of the electron.

Answer to Problem 51PQ

The rest mass energy of an electron is $512\text{keV}$ or $8.20\times {10}^{-14}\text{J}$.

Explanation of Solution

Write the expression for the rest mass energy.

    $E_{\text{rest}}=m_{\text{rest}}{c}^2$                        (I)

Here, $m_{\text{rest}}$ is the rest mass,$c$ is the speed of the light and $E_{\text{rest}}$ is the total energy of the electron at rest.

Conclusion:

Substitute $9.11\times {10}^{-31}\text{kg}$ for $m_{\text{rest}}$ and $3.00\times {10}^8\text{m}/\text{s}$ for $c$ in equation (I).

    $\begin{array}{c}{E}_{\text{rest}}=\left(9.11\times {10}^{-31}\text{kg}\right){\left(3.00\times {10}^8\text{m}/\text{s}\right)}^2\\ =8.20\times {10}^{-14}\text{J}\end{array}$                 (II)

Convert $\text{J}$ to $\text{keV}$ by multiplying with $\left(\frac{1\text{keV}}{1.602\times {10}^{-16}\text{J}}\right)$ in equation (II).

Here, $\text{keV}$ is the kilo electron volt, and $\text{J}$ is joule.

    $\begin{array}{c}{E}_{\text{rest}}=8.20\times {10}^{-14}\text{J}\left(\frac{1\text{keV}}{1.602\times {10}^{-16}\text{J}}\right)\\ =\frac{8.20\times {10}^{-14}}{1.602\times {10}^{-16}}\left(\frac{\text{keV}\times \text{J}}{\text{J}}\right)\\ =512\text{keV}\end{array}$

Therefore, the rest mass energy of an electron is $512\text{keV}$ or $8.20\times {10}^{-14}\text{J}$.

(b)

To determine

The total energy of an electron.

Answer to Problem 51PQ

The total energy of an electron is $1.30\times {10}^{-13}\text{J}$ or $\text{812}\text{keV}$.

Explanation of Solution

Write the expression for total energy of an electron.

    $E=K+E_{\text{rest}}$                  (III)

Here, $E$ is the total energy of the electron, $K$ is the kinetic energy and $E_{\text{rest}}$ is the total energy of the electron at rest.

Conclusion:

Substitute $300.0\text{keV}$ for $K$ and $512\text{keV}$ for $E_{\text{rest}}$ in equation (III).

    $\begin{array}{c}E=300.0\text{keV}+512\text{keV}\\ =\text{812}\text{keV}\end{array}$                     (IV)

Convert $\text{keV}$ to $\text{J}$ by multiplying with $\left(\frac{1.602\times {10}^{-16}\text{J}}{1\text{keV}}\right)$ in equation (IV).

    $\begin{array}{c}E\text{=812}\text{keV}\left(\frac{1.602\times {10}^{-16}\text{J}}{1\text{keV}}\right)\\ =\text{812}\left(1.602\times {10}^{-16}\right)\left(\frac{\text{keV}\times \text{J}}{\text{keV}}\right)\\ =1.30\times {10}^{-13}\text{J}\end{array}$

Therefore, the total energy of an electron is $1.30\times {10}^{-13}\text{J}$ or $\text{812}\text{keV}$.

(c)

To determine

The speed of the electron.

Answer to Problem 51PQ

The speed of the electron is $0.7762c$.

Explanation of Solution

Write the expression of the relativistic kinetic energy of the particle.

    $K\left(\gamma -1\right)m_{\text{rest}}{c}^2$                                          (V)

Here, $m_{\text{rest}}$ is the rest mass of the particle, $v$ is the moving velocity of the particle velocity, $\gamma =\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}$ is the Lorentz factor, $c$ is the speed of light and $K$ is the relativistic energy.

Write the expression for Lorentz factor.

    $\gamma =\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}$

Substitute $\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}$ for $\gamma$ in the Equation (V)

    $K=\left(\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}-1\right)m_{\text{rest}}{c}^2$                               (VI)

Convert the unit of $K$ from $\text{keV}$ to $\text{J}$ by multiplying $\left(\frac{1.602\times {10}^{-16}\text{J}}{1\text{keV}}\right)$ with $300.0\text{keV}$.

    $\begin{array}{c}K=300.0\text{keV}\\ \text{=}300.0\text{keV}\left(\frac{1.602\times {10}^{-16}\text{J}}{1\text{keV}}\right)\\ =4.806\times {10}^{-14}\text{J}\end{array}$                             (VII)

Conclusion:

Substitute $4.806\times {10}^{-14}\text{J}$ for $K$, $9.11\times {10}^{-31}\text{kg}$ for $m_{\text{rest}}$ and $3.00\times {10}^8\text{m}/\text{s}$ for $c$ in equation (VI).

    $\begin{array}{c}4.806\times {10}^{-14}\text{J}=\left(\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}-1\right)\left(9.11\times {10}^{-31}\text{kg}\right){\left(3.00\times {10}^8\text{m}/\text{s}\right)}^2\\ \frac{4.806\times {10}^{-14}\text{J}}{\left(9.11\times {10}^{-31}\text{kg}\right){\left(3.00\times {10}^8\text{m}/\text{s}\right)}^2}=\left(\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}-1\right)\\ \sqrt{1-{\left(\frac{v}{c}\right)}^2}=\frac{1}{1+\left(\frac{4.806\times {10}^{-14}\text{J}}{\left(9.11\times {10}^{-31}\text{kg}\right){\left(3.00\times {10}^8\text{m}/\text{s}\right)}^2}\right)}\\ 1-{\left(\frac{v}{c}\right)}^2={\left(\frac{1}{1+0.586}\right)}^2\end{array}$

Solve the equation further to find $v$.

    $\begin{array}{c}1-{\left(\frac{v}{c}\right)}^2=0.3975\\ {\left(\frac{v}{c}\right)}^2=\left(1-0.3975\right)\\ \left(\frac{v}{c}\right)=\sqrt{0.6025}\\ =0.7762c\end{array}$

Therefore, the speed of the electron is $0.7762c$.