Chapter 39, Problem 66AP

(a)

To determine

The speed of a proton that has the same kinetic energy as the electron.

Answer to Problem 66AP

The speed of the proton that has the same kinetic energy as the electron is $\underset{\_}{0.0236c}$.

Explanation of Solution

Write the expression for the kinetic energy of the electron.

  $K_e=m_e{c}^2\left({\gamma }_e-1\right)$                                                                                                      (I)

Here, $K_e$ is the kinetic energy of the electron, $m_e$ is the mass of the electron, $c$ is the speed of light, and ${\gamma }_e$ is the Lorentz factor for the electron.

Write the expression for the kinetic energy of the proton.

  $K_p=m_p{c}^2\left({\gamma }_p-1\right)$                                                                                                    (II)

Here, $K_p$ is the kinetic energy of the proton, $m_p$ is the mass of the proton, and ${\gamma }_p$ is the Lorentz factor of the proton.

Given that the kinetic energy of electron and proton are equal.

  $K_p=K_e$                                                                                                                   (III)

Use expressions (I) and (II) in expression (III).

  $m_e{c}^2\left({\gamma }_e-1\right)=m_p{c}^2\left({\gamma }_p-1\right)$                                                                                    (IV)

Solve expression (IV) for ${\gamma }_p$.

  ${\gamma }_p=1+\frac{m_e{c}^2\left({\gamma }_e-1\right)}{m_p{c}^2}$                                                                                                (V)

Write the expression for Lorentz factor of electron.

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

Here, $u_e$ is the speed of electron.

Write the expression for Lorentz factor of proton.

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

Here, $u_p$ is the speed of proton.

Use expression (VII) for $u_p$.

  $u_p=c\sqrt{1-{\gamma }_p^{-2}}$                                                                                                      (VIII)

Conclusion:

Substitute $0.750c$ for $u_e$ in equation (VI) to find ${\gamma }_e$.

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

Substitute $9.1\times {10}^{-31}\text{kg}$ for $m_e$, $3.00\times {10}^8\text{m}/\text{s}$ for $c$, $1.67\times {10}^{-27}\text{kg}$ for $m_p$, and $1.5119$ for ${\gamma }_e$ in equation (V) to find ${\gamma }_p$.

  $\begin{array}{c}{\gamma }_p=1+\frac{\frac{\left(9.1\times {10}^{-31}\text{kg}\right){\left(3.00\times {10}^8\text{m}/\text{s}\right)}^2\left(1.5119-1\right)}{1.6\times {10}^{-19}{\text{kgm}}^2/{\text{s}}^{\text{2}}}}{\frac{\left(1.67\times {10}^{-27}\text{kg}\right){\left(3.00\times {10}^8\text{m}/\text{s}\right)}^2}{1.6\times {10}^{-19}{\text{kgm}}^2/{\text{s}}^{\text{2}}}}\\ =1.000279\end{array}$

Substitute $1.000279$ for ${\gamma }_p$ in equation (VIII) to find $u_p$.

  $\begin{array}{c}{u}_p=c\sqrt{1-{\left(1.000279\right)}^{-2}}\\ =0.0236c\end{array}$

Therefore, the speed of the proton that has the same kinetic energy as the electron is $\underset{\_}{0.0236c}$.

(b)

To determine

The speed of the proton that has the same momentum as the electron.

Answer to Problem 66AP

The speed of the proton that has the same momentum as the electron is $\underset{\_}{6.18\times {10}^{-4}c}$.

Explanation of Solution

Write the expression for the relativistic momentum of electron.

  $p_e={\gamma }_e{m}_e{u}_e$                                                                                                             (IX)

Here, $p_e$ is the relativistic momentum of electron.

Write the expression for the relativistic momentum of proton.

  $p_p={\gamma }_p{m}_p{u}_p$                                                                                                             (X)

Here, $p_p$ is the relativistic momentum of proton.

Momentum of electron and proton are equal.

  $p_e=p_p$                                                                                                                   (XI)

Use expressions (IX) and (X) in (XI).

  ${\gamma }_e{m}_e{u}_e={\gamma }_p{m}_p{u}_p$                                                                                                    (XII)

Solve equation (XII) for ${\gamma }_p$.

  $u_p=\frac{{\gamma }_e{m}_e{u}_e}{m_p{\gamma }_p}$                                                                                                          (XIII)

Conclusion:

Substitute $1.5119$ for ${\gamma }_e$, $9.1\times {10}^{-31}\text{kg}$ for $m_e$, $0.750c$ for $u_e$, $1.67\times {10}^{-27}\text{kg}$ for $m_p$, and $1.000279$ for ${\gamma }_p$ in equation (XIII) to find $u_p$.

  $\begin{array}{c}{u}_p=\frac{\left(1.5119\right)\left(9.1\times {10}^{-31}\text{kg}\right)\left(0.750c\right)}{\left(1.67\times {10}^{-27}\text{kg}\right)\left(1.000279\right)}\\ =6.18\times {10}^{-4}c\end{array}$

Therefore, the speed of the proton that has the same momentum as the electron is $\underset{\_}{6.18\times {10}^{-4}c}$.