Chapter 26, Problem 55P

(a)

To determine

The kinetic energy of ion.

Answer to Problem 55P

The kinetic energy of ion is $\underset{\_}{750\text{MeV}}$.

Explanation of Solution

Write the expression for kinetic energy

  $K=qV$                                                                                                                     (I)

Here, $K$ is the kinetic energy, $q$ is the charge, and $V$ is the potential

Conclusion:

Substitute, $+6e$ for $q$, $125\text{MV}$ for $V$ in equation (I)

  $\begin{array}{c}K=+6\text{e}\times 125\text{MV}\\ \text{=750}\text{MeV}\end{array}$

Therefore, the kinetic energy of ion is $\underset{\_}{750\text{MeV}}$.

(b)

To determine

The speed of ions as measured in lab frame.

Answer to Problem 55P

The speed of ions as measured in lab frame is $\underset{\_}{0.34908c}$.

Explanation of Solution

Write the expression for relative kinetic energy

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

Here, $\gamma$ is the Lorentz factor, $m$ is the mass, and $c$ is the speed of light

Substitute $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ for $\gamma$ in equation (II), and Rearrange to obtain an expression for $v$

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

Conclusion:

Substitute, $750\text{MeV}$ for $K$, $11172\text{MeV}$ for $m{c}^2$ in equation (III)

  $\begin{array}{c}v=c\sqrt{1-{\left(1+\frac{750\text{MeV}}{750\text{MeV}}\right)}^{-2}}\\ =0.34908c\end{array}$

Therefore, the speed of ions as measured in lab frame is $\underset{\_}{0.34908c}$.

(c)

To determine

The length of accelerator in reference frame.

Answer to Problem 55P

The length of accelerator in reference frame is $\underset{\_}{7.03\text{m}}$.

Explanation of Solution

Write the expression for length contraction

  $L=\frac{L_0}{\gamma }$                                                                                                                    (IV)

Here, $L$ is the relative length of accelerator, $L_0$ is the actual length

Substitute, $1+\frac{K}{m{c}^2}$ for $\gamma$ in equation (IV)

  $L=\frac{L_0}{1+\frac{K}{m{c}^2}}$                                                                                                            (V)

Conclusion:

Substitute, $11172\text{MeV}$ for $m{c}^2$, $750\text{MeV}$ for $K$, and $7.50\text{m}$ for $L$ in equation (V)

  $\begin{array}{c}L=\frac{7.50\text{m}}{\left(1+\frac{750\text{MeV}}{11172\text{MeV}}\right)}\\ =7.03\text{m}\end{array}$

Therefore, the length of accelerator in reference frame is $\underset{\_}{7.03\text{m}}$.