Chapter 30, Problem 28P

(a)

To determine

Find the speed of proton.

Answer to Problem 28P

The speed of proton is $0.999999991c$.

Explanation of Solution

The proton is accelerated energy is $7\text{ TeV}$.

Write the equation for relativistic energy.

$E=\frac{m{c}^2}{\sqrt{1-\frac{v^2}{c^2}}}$

Here, $E$ is the energy proton, $m$ is the mass of proton, $v$ is speed of proton, and $c$ is the speed of light.

Rearrange the above equation to get the $v$.

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

Conclusion:

Substitute $7\text{ TeV}$ for $E$ and $398.272\text{ MeV}$ for $m{c}^2$ in equation I.

$\begin{array}{c}v=c\sqrt{1-\frac{{\left(398.272\text{ MeV}\right)}^2}{{\left(7\text{ TeV}\right)}^2}}\\ =0.999999991c\end{array}$

Therefore, the speed of proton is $0.999999991c$.

(b)

To determine

Find time it would take proton to make one rotation.

Answer to Problem 28P

The time taken by proton to make one rotation is $12\text{ ns}$.

Explanation of Solution

The circumference of tunnel is $27\text{ km}$.

Write the relativistic equation to find time take by proton.

$t=\frac{L_0}{c}\sqrt{1-\frac{v^2}{c^2}}$ (II)

Here, the $t$ is the time taken by proton and $L_0$ is the circumference of the tunnel.

Conclusion:

Substitute $27\text{ km}$ for $L_0$, $3\times {10}^8\text{ m/s}$ for $c$  and $0.999999991c$ for $v$ in equation II.

$\begin{array}{c}t=\frac{27\text{ km}}{3\times {10}^8\text{ m/s}}\sqrt{1-\frac{{\left(0.999999991c\right)}^2}{c^2}}\\ =\frac{27\times {10}^3\text{ m}}{3\times {10}^8\text{ m/s}}\left(1.3416\times {10}^{-4}\right)\\ =0.0012\times {10}^{-5}\text{ s}\\ \text{=12 ns}\end{array}$

Therefore, the time taken by proton to take on rotation is $12\text{ ns}$.

(c)

To determine

Find time it would take proton to make one rotation in the reference frame.

Answer to Problem 28P

The time taken by proton to make one rotation in reference frame is $90\text{ μs}$.

Explanation of Solution

Write the equation for time taken by proton in reference frame.

$t_0=\frac{L_0}{c}$ (III)

Here, $t_0$ is the time taken by proton to take one rotation in reference frame.

Conclusion:

Substitute $27\text{ km}$ for $L_0$ and $3\times {10}^8\text{ m/s}$ for $c$ in equation III.

$\begin{array}{c}{t}_0=\frac{27\text{ km}}{3\times {10}^8\text{ m/s}}\\ =90\text{μs}\end{array}$

Therefore, the time taken by proton to make one rotation in reference frame is $90\text{ μs}$.

(d)

To determine

Find the de Broglie wavelength of proton.

Answer to Problem 28P

The de Broglie wavelength of proton is $1.8\times {10}^{-19}\text{ m}$.

Explanation of Solution

Write the relativistic de Broglie equation.

$\lambda =\frac{h}{mv}\sqrt{1-\frac{v^2}{c^2}}$ (IV)

Here, $\lambda$ is de Broglie wavelength and $h$  is the Plank’s constant.

Conclusion:

Substitute $0.999999991c$ for $v$, $6.626\times {10}^{-34}\text{ Js}$ for $h$, $1.673\times {10}^{-27}\text{ kg}$ for m and $3\times {10}^8\text{ m/s}$ for $c$ in equation IV.

$\begin{array}{c}\lambda =\frac{6.626\times {10}^{-34}\text{ Js}}{\left(1.673\times {10}^{-27}\text{ kg}\right)\left(0.999999991\right)\left(3\times {10}^8\text{ m/s}\right)}\sqrt{1-\frac{{\left(0.999999991c\right)}^2}{c^2}}\\ =1.8\times {10}^{-19}\text{ m}\end{array}$

Therefore, the de Broglie wavelength of proton is $1.8\times {10}^{-19}\text{ m}$.

(e)

To determine

Find the mosquito’s speed.

Answer to Problem 28P

The mosquito’s speed is $1.5\text{ m/s}$.

Explanation of Solution

First the decision has to be made whether to use relativistic or nonrelativistic formula.

Write the kinetic energy in joules,

$7\text{ TeV}=\left(7\times {10}^{12}\text{ eV}\right)\left(1.602\times {10}^{-19}\text{ J/eV}\right)=1.1\text{ μJ}$

The rest energy of mosquito is given below,

$E_0=m{c}^2=\left(1.0\times {10}^{-6}\text{ kg}\right){\left(3.0\times {10}^8\text{ m/s}\right)}^2=9.0\times {10}^{10}\text{ J}$

Therefore, the rest energy of mosquito is much higher than the kinetic energy, thus the nonrelativistic equation must be used to calculate the speed.

Write the nonrelativistic equation for speed.

$K=\frac{1}{2}m{v}^2$

Here, $K$  is the kinetic energy, $m$  is the mass of mosquito and $v$ is speed of mosquito.

Rewrite the above equation to find speed.

$v=\sqrt{\frac{2K}{m}}$ (V)

Here, $\lambda$ is de Broglie wavelength and $h$  is the Plank’s constant.

Conclusion:

Substitute $1.1\text{ μJ}$ for $K$ and $1.0\times {10}^{-6}\text{ kg}$ for $m$ in equation V.

$\begin{array}{c}v=\sqrt{\frac{2\left(1.1\text{ μJ}\right)}{1.0\times {10}^{-6}\text{ kg}}}\\ =\sqrt{\frac{2.2\times {10}^{-6}{\text{ kgm}}^2/s^2}{1.0\times {10}^{-6}\text{ kg}}}\\ =1.5\text{ m/s}\end{array}$

Therefore, the mosquito’s speed is $1.5\text{ m/s}$.