Chapter 39, Problem 50PQ

(a)

To determine

The observed mass of electron in the lab.

Answer to Problem 50PQ

The observed mass of electron is $1.444\times {10}^{-30}\text{kg}$.

Explanation of Solution

Write relativistic kinetic energy equation of an electron.

    $K=m{c}^2-m_{\text{rest}}{c}^2$                                                             (I)

Here, $K$ is the kinetic energy, $m$ is the observed mass of electron, $m_{\text{rest}}$ is the rest mass of electron and $c$ is the speed of light.

Write the equation of mass transformation.

    $m=\gamma m_{\text{rest}}$                                                             (II)

Here, $m_{\text{rest}}$ is the rest mass of baseball and $\gamma$ is the Lorentz factor.

Write the equation of Lorentz factor.

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

Here, $v$ is the speed of electron.

Substitute, $\gamma =\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}$ in the equation (II).

    $m=\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}{m}_{\text{rest}}$                                          (III)

Substitute $\gamma m_{\text{rest}}$ for $m$ in equation (I).

    $\begin{array}{c}K=\gamma m_{\text{rest}}{c}^2-m_{\text{rest}}{c}^2\\ =\left(\gamma -1\right)m_{\text{rest}}{c}^2\end{array}$                              (IV)

Substitute, $\gamma =\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}$ in the above equation.

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

Write the equation for relativistic momentum of equation.

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

Here, $p$ is the momentum of electron.

Convert the unit of $K$ from $\text{keV}$ to $\text{J}$.

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

Substitute $4.806\times {10}^{-14}\text{J}$ for $K$, $9.11\times {10}^{-31}\text{kg}$ for $m_{\text{rest}}$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in equation (V).

    $\begin{array}{c}4.806\times {10}^{-14}\text{J}=\left(\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}-1\right)\left(9.11\times {10}^{-31}\text{kg}\right){\left(3\times {10}^8\text{m}/\text{s}\right)}^2\\ \frac{4.806\times {10}^{-14}\text{J}}{\left(9.11\times {10}^{-31}\text{kg}\right)\times \left(9\times {10}^{16}{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right)}=\left(\frac{1}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}-1\right)\end{array}$

Rearrange the above equation.

    $\begin{array}{c}\sqrt{1-{\left(\frac{v}{c}\right)}^2}=\frac{1}{1+\left(\frac{4.806\times {10}^{-14}\text{J}}{\left(9.11\times {10}^{-31}\text{kg}\right)\times \left(9\times {10}^{16}{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\right)}\right)}\\ 1-{\left(\frac{v}{c}\right)}^2=\left(\frac{1}{1+0.586\text{J}\cdot 1/\text{kg}\cdot 1/{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\\ {\left(\frac{v}{c}\right)}^2=1-{\left(0.63\right)}^2\\ v^2=0.602c^2\end{array}$

Take square roots on both sides.

    $v=0.776c$

Conclusion:

Substitute $0.776c$ for $v$, $9.11\times {10}^{-31}\text{kg}$ for $m_{\text{rest}}$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in equation (III).

    $\begin{array}{c}m=\frac{1}{\sqrt{1-{\left(\frac{0.776c}{c}\right)}^2}}\left(9.11\times {10}^{-31}\text{kg}\right)\\ =\frac{1}{\sqrt{1-{\left(\frac{0.776\times 3\times {10}^8\text{m}/\text{s}}{3\times {10}^8\text{m}/\text{s}}\right)}^2}}\left(9.11\times {10}^{-31}\text{kg}\right)\\ =\text{1}\text{.58545}\left(9.11\times {10}^{-31}\text{kg}\right)\\ =1.444\times {10}^{-30}\text{kg}\end{array}$

Therefore, the observed mass of electron is $1.444\times {10}^{-30}\text{kg}$.

(b)

To determine

The momentum of electron.

Answer to Problem 50PQ

The momentum of electron is $3.362\times {10}^{-22}\text{kg}\cdot \text{m}/\text{s}$.

Explanation of Solution

Substitute $0.776c$ for $v$, $9.11\times {10}^{-31}\text{kg}$ for $m_{\text{rest}}$ and $3\times {10}^8\text{m}/\text{s}$ for $c$ in equation (VI).

    $\begin{array}{c}p=\frac{1}{\sqrt{1-{\left(\frac{0.776c}{c}\right)}^2}}\left(9.11\times {10}^{-31}\text{kg}\right)\left(0.776c\right)\\ =\frac{1}{\sqrt{1-{\left(\frac{\left(0.776\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{3\times {10}^8\text{m}/\text{s}}\right)}^2}}\left(9.11\times {10}^{-31}\text{kg}\right)\left(0.776\times 3\times {10}^8\text{m}/\text{s}\right)\\ =1.58545\left(9.11\times {10}^{-31}\text{kg}\right)\left(0.776\times 3\times {10}^8\text{m}/\text{s}\right)\\ =3.362\times {10}^{-22}\text{kg}\cdot \text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the momentum of electron is $3.362\times {10}^{-22}\text{kg}\cdot \text{m}/\text{s}$.