Chapter 26, Problem 51P

(a)

To determine

The speed of the electrons produced by the electron accelerator.

Answer to Problem 51P

The speed of the electrons produced by the electron accelerator is $\underset{\_}{0.99980c}$.

Explanation of Solution

Given that the kinetic energy of the electrons is $25\text{MeV}$, the rest mass energy of electron is $0.5110\text{MeV}$.

Write the expression for relativistic kinetic energy.

    $K=\left(\gamma -1\right)m{c}^2$                                                                                                         (I)

Here, $K$ is the kinetic energy of the electrons, $\gamma$ is the Lorentz factor, $m$ is the mass of the particle, $c$ is the speed of light.

Write the expression for the Lorentz factor.

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

Here, $v$ is the speed of the spaceship.

Use equation (II) in equation (I).

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

Solve equation (III) for $v$.

    $\begin{array}{c}1+\frac{K}{m{c}^2}=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\\ 1-\frac{v^2}{c^2}={\left(1+\frac{K}{m{c}^2}\right)}^{-2}\\ v=c\sqrt{1-{\left(1+\frac{K}{m{c}^2}\right)}^{-2}}\end{array}$                                                                                  (IV)

Conclusion:

Substitute $25\text{MeV}$ for $K$, $0.5110\text{MeV}$ for $m{c}^2$ in equation (IV) to find $v$.

    $\begin{array}{c}v=c\sqrt{1-{\left(1+\frac{25\text{MeV}}{0.5110\text{MeV}}\right)}^{-2}}\\ =0.99980c\end{array}$

Therefore, the speed of the electrons produced by an $\underset{\_}{0.99980c}$.

(b)

To determine

The time taken for the electrons to travel the distance in the reference frame of electrons.

Answer to Problem 51P

The time taken for the electrons to travel the distance in the reference frame of electrons is $\underset{\_}{0.010\text{ns}}$.

Explanation of Solution

Write the expression for the time taken in the reference frame of electrons.

    $\Delta t_0=\frac{\Delta t}{\gamma }$                                                                                                                 (V)

Here, $\Delta t_0$ is the time interval in the reference frame of the electrons, $\Delta t$ is the time taken by the electrons to exit the accelerator and reach the patient in the reference frame.

Write the expression for the time taken by the electrons to exit the accelerator and reach the patient in the reference frame.

    $\Delta t=\frac{\Delta x}{v}$                                                                                                                (VI)

Here, $\Delta x$ is the distance of the accelerator from the patient, $v$ is the speed of the electrons.

Use equation (II) and (VI) in equation (V),

    $\Delta t_0=\frac{\Delta x\sqrt{1-\frac{v^2}{c^2}}}{v}$                                                                                                  (VII)

Conclusion:

Substitute $15\text{cm}$ for $\Delta x$, $0.99980c$ for $v$, and $2.998\times {10}^8\text{m}/\text{s}$ for $c$ in equation (VII) to find $\Delta t_0$.

    $\begin{array}{c}\Delta t_0=\frac{15.0\text{cm}\times \sqrt{1-\frac{{\left(0.99980c\right)}^2}{c^2}}}{0.99980\left(2.998\times {10}^8\text{m}/\text{s}\right)}\\ =\frac{\left(15.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\sqrt{1-\frac{{\left(0.99980c\right)}^2}{c^2}}}{0.99980\left(2.998\times {10}^8\text{m}/\text{s}\right)}\\ =0.010\text{ns}\end{array}$

Therefore, the time interval in the reference frame of electrons is $\underset{\_}{0.010\text{ns}}$.