Chapter 39, Problem 39.66AP

(a)

To determine

The speed of a proton when its kinetic energy is equal to the kinetic energy of the electron.

Answer to Problem 39.66AP

The speed of a proton is $0.0236c$.

Explanation of Solution

Given info: The speed of an electron is $0.750c$.

Write the equation of kinetic energy of proton.

$K_{\text{p}}=\left({\gamma }_{\text{p}}-1\right)m_{\text{p}}{c}^2$

Here,

${\gamma }_{\text{p}}$ is a constant.

$m_{\text{p}}$ is the mass of proton.

$c$ is the speed of light.

The value of ${\gamma }_{\text{p}}$ is $\frac{1}{\sqrt{1-{\left(u_{\text{p}}/c\right)}^2}}$, where $u_{\text{p}}$ is the speed of a proton.

Substitute $\frac{1}{\sqrt{1-{\left(u_{\text{p}}/c\right)}^2}}$ for ${\gamma }_{\text{p}}$ in above equation to find $K_{\text{p}}$.

$K_{\text{p}}=\left(\frac{1}{\sqrt{1-{\left(u_{\text{p}}/c\right)}^2}}-1\right)m_{\text{p}}{c}^2$

Write the equation of kinetic energy of electron.

$K_{\text{e}}=\left({\gamma }_{\text{e}}-1\right)m_{\text{e}}{c}^2$

Here,

${\gamma }_{\text{e}}$ is a constant.

$m_{\text{e}}$ is the mass of an electron.

The value of ${\gamma }_{\text{e}}$ is,

${\gamma }_{\text{e}}=\frac{1}{\sqrt{1-{\left(u_{\text{e}}/c\right)}^2}}$

Here,

$u_{\text{e}}$ is the speed of an electron.

Substitute $0.750c$ for $u_{\text{e}}$ in above equation to find ${\gamma }_{\text{e}}$.

$\begin{array}{c}{\gamma }_{\text{e}}=\frac{1}{\sqrt{1-{\left(0.750c/c\right)}^2}}\\ {\gamma }_{\text{e}}=1.5119\end{array}$

Thus, the value of ${\gamma }_{\text{e}}$ is $1.5119$.

Substitute $1.5119$ for ${\gamma }_{\text{e}}$ in above equation to find $K_{\text{e}}$.

$\begin{array}{c}{K}_{\text{e}}=\left(1.5119-1\right)m_{\text{e}}{c}^2\\ K_{\text{e}}=0.5119m_{\text{e}}{c}^2\end{array}$

The kinetic energy of proton is equal to the kinetic energy of electron.

$K_{\text{p}}=K_{\text{e}}$

Substitute $\left(\frac{1}{\sqrt{1-{\left(u_{\text{p}}/c\right)}^2}}-1\right)m_{\text{p}}{c}^2$ for $K_{\text{p}}$ and $0.5119m_{\text{e}}{c}^2$ for $K_{\text{e}}$ in above equation.

$\begin{array}{c}\left(\frac{1}{\sqrt{1-{\left(u_{\text{p}}/c\right)}^2}}-1\right)m_{\text{p}}{c}^2=0.5119m_{\text{e}}{c}^2\\ \frac{1}{\sqrt{1-{\left(u_{\text{p}}/c\right)}^2}}=\frac{0.5119m_{\text{e}}}{m_{\text{p}}}+1\\ u_{\text{p}}=c\sqrt{1-\frac{1}{{\left(\frac{0.5119m_{\text{e}}}{m_{\text{p}}}+1\right)}^2}}\end{array}$

The mass of proton is $1.67\times {10}^{-27}\text{kg}$ and the mass of an electron is $9.11\times {10}^{-31}\text{kg}$.

Substitute $1.67\times {10}^{-27}\text{kg}$ for $m_{\text{p}}$ and $9.11\times {10}^{-31}\text{kg}$ for $m_{\text{e}}$ in above equation to find $u_{\text{p}}$.

$\begin{array}{c}{u}_{\text{p}}=c\sqrt{1-\frac{1}{{\left(\frac{0.5119\left(9.11\times {10}^{-31}\text{kg}\right)}{\left(1.67\times {10}^{-27}\text{kg}\right)}+1\right)}^2}}\\ =0.0236c\end{array}$

Conclusion:

Therefore, the speed of a proton is $0.0236c$.

(b)

To determine

The speed of a proton when its momentum is equal to the momentum of an electron.

Answer to Problem 39.66AP

The speed of a proton is $6.18\times {10}^{-4}c$.

Explanation of Solution

Given info: The speed of an electron is $0.750c$.

Write the equation of momentum of proton.

$p_{\text{p}}={\gamma }_{\text{p}}{m}_{\text{p}}{u}_{\text{p}}$

Substitute $\frac{1}{\sqrt{1-{\left(u_{\text{p}}/c\right)}^2}}$ for ${\gamma }_{\text{p}}$ in above equation to find $p_{\text{p}}$.

$p_{\text{p}}=\frac{1}{\sqrt{1-{\left(u_{\text{p}}/c\right)}^2}}{m}_{\text{p}}{u}_{\text{p}}$

Write the equation of momentum of electron.

$p_{\text{e}}={\gamma }_{\text{e}}{m}_{\text{e}}{u}_{\text{e}}$

Substitute $1.5119$ for ${\gamma }_{\text{e}}$, $9.11\times {10}^{-31}\text{kg}$ for $m_{\text{e}}$ and $0.750c$ for $u_{\text{e}}$ in above equation to find $p_{\text{e}}$.

$\begin{array}{c}{p}_{\text{e}}=\left(1.5119\right)\left(9.11\times {10}^{-31}\text{kg}\right)\left(0.750c\right)\\ p_{\text{e}}=\left(1.033\times {10}^{-30}\text{kg}\right)c\end{array}$

The momentum of proton and electron are same.

$p_{\text{p}}=p_{\text{e}}$

Substitute $\frac{1}{\sqrt{1-{\left(u_{\text{p}}/c\right)}^2}}{m}_{\text{p}}{u}_{\text{p}}$ for $p_{\text{p}}$ and $\left(1.033\times {10}^{-30}\text{kg}\right)c$ for $p_{\text{e}}$ in above equation.

$\frac{1}{\sqrt{1-{\left(u_{\text{p}}/c\right)}^2}}{m}_{\text{p}}{u}_{\text{p}}=\left(1.033\times {10}^{-30}\text{kg}\right)c$

Substitute $1.67\times {10}^{-27}\text{kg}$ for $m_{\text{p}}$ in above equation to find $u_{\text{p}}$.

$\begin{array}{c}\frac{1}{\sqrt{1-{\left(u_{\text{p}}/c\right)}^2}}\left(1.67\times {10}^{-27}\text{kg}\right)u_{\text{p}}=\left(1.033\times {10}^{-30}\text{kg}\right)c\\ \frac{u_{\text{p}}}{c}=\left(6.18\times {10}^{-4}\right)\sqrt{1-{\left(\frac{u_{\text{p}}}{c}\right)}^2}\\ u_{\text{p}}=c\sqrt{\frac{3.819\times {10}^{-7}}{1+\left(3.819\times {10}^{-7}\right)}}\\ u_{\text{p}}=6.18\times {10}^{-4}c\end{array}$

Conclusion:

Therefore, the speed of a proton is $6.18\times {10}^{-4}c$.