Chapter 41, Problem 36SP

Find the speed and momentum of a proton $\left(m=1.67\times {10}^{-27}\text{ kg}\right)$ that has been accelerated through a potential difference of 2000 MV. (We call this a 2 GeV proton.) Give your answers to three significant figures.

To determine

The speed and momentum of a protonthat is accelerated through the potential difference of 2000 MVif the mass of proton is $1.67\times {10}^{-27}\text{ kg}$.

Answer to Problem 36SP

Solution:

$0.948c$ and $1.49\times {10}^{-18}\text{ kg}\cdot \text{m/s}$

Explanation of Solution

Given data:

Mass of proton is $1.67\times {10}^{-27}\text{ kg}$.

Potential difference through which the proton is accelerated is $2000\text{ MV}$.

Formula used:

Write the expression for the potential energy of a charge due to a potential difference:

$PE=qV$

Here, $V$ is the potential difference and $q$ is the charge of the particle.

Write the expression for kinetic energy in case of relativistic motion:

$KE=\sqrt{m^2{c}^4+p^2{c}^2}-m{c}^2$

Here, $c$ is the speed of light, $KE$ is the kinetic energy, $p$ is the momentum, and $m$ is the mass of proton.

Write the expression for relativistic momentum:

$p=\gamma mv$

Write the expression for $\gamma$

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

Here, $c$ is the speed of light and $v$ is the speed or velocity.

Explanation:

Recall the expression for potential energy of a charged particle due potential difference.

$PE=qV$

The potential difference will cause an acceleration to the charge particle. Therefore, it will result in kinetic energy of the charged particle. Hence,

$KE=PE$

Substitute $qV$ for $KE$

$KE=qV$

Substitute $2000\text{ MV}$ for $V$ and $e$ for $q$

$KE=e\left(2000\text{ MV}\right)$

Substitute $1.6\times {10}^{-19}\text{ C}$ for $e$

$\begin{array}{c}KE=\left(1.6\times {10}^{-19}\text{ C}\right)\left(2000\text{ MV}\right)\\ =\left(1.6\times {10}^{-19}\text{ C}\right)\left(2000\text{ MV}\right)\left(\frac{{10}^6\text{ V}}{1\text{ MV}}\right)\\ =3.2\times {10}^{-10}\text{ J}\end{array}$

Recall the expression for kinetic energy for the relativistic motion of proton:

$KE=\sqrt{m^2{c}^4+p^2{c}^2}-m{c}^2$

Substitute $3.2\times {10}^{-10}\text{ J}$ for $KE$, $1.67\times {10}^{-27}\text{ kg}$ for $m$, and $2.998\times {10}^8\text{ m/s}$ for $c$

$\begin{array}{c}3.2\times {10}^{-10}\text{ J}=\left\{\begin{array}{l}\sqrt{{\left(1.67\times {10}^{-27}\text{ kg}\right)}^2{\left(2.998\times {10}^8\text{ m/s}\right)}^4+p^2{\left(2.998\times {10}^8\text{ m/s}\right)}^2}\\ -\left(1.67\times {10}^{-27}\text{ kg}\right){\left(2.998\times {10}^8\text{ m/s}\right)}^2\end{array}\right\}\\ 3.2\times {10}^{-10}\text{ J}=\sqrt{2.25\times {10}^{-20}{\text{ J}}^2+p^2\left(8.99\times {10}^{16}{\text{ m}}^{\text{2}}{\text{/s}}^{\text{2}}\right)}-\text{1}{\text{.50×10}}^{-\text{10}}\text{ J}\end{array}$

Further solve

$\begin{array}{c}3.2\times {10}^{-10}\text{ J}+\text{1}{\text{.50×10}}^{-\text{10}}\text{ J}=\sqrt{2.25\times {10}^{-20}{\text{ J}}^2+p^2\left(8.99\times {10}^{16}{\text{ m}}^{\text{2}}{\text{/s}}^{\text{2}}\right)}\\ 4.7\times {10}^{-10}\text{ J}=\sqrt{2.25\times {10}^{-20}{\text{ J}}^2+p^2\left(8.99\times {10}^{16}{\text{ m}}^{\text{2}}{\text{/s}}^{\text{2}}\right)}\\ 22.09\times {10}^{-20}{\text{ J}}^2=2.25\times {10}^{-20}{\text{ J}}^2+p^2\left(8.99\times {10}^{16}{\text{ m}}^{\text{2}}{\text{/s}}^{\text{2}}\right)\\ p^2\left(8.99\times {10}^{16}{\text{ m}}^{\text{2}}{\text{/s}}^{\text{2}}\right)=19.84\times {10}^{-20}{\text{ J}}^2\\ p^2=2.206\times {10}^{-36}\\ p=1.49\times {10}^{-18}\text{ kg}\cdot \text{m/s}\end{array}$

Recall the expression for $\gamma$

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

Recall the expression for relativistic momentum of the proton:

$p=\gamma mv$

Substitute $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ for $\gamma$

$p=\left(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\right)mv$

Substitute $1.49\times {10}^{-18}\text{ kg}\cdot \text{m/s}$ for $p$ and $1.67\times {10}^{-27}\text{ kg}$ for $m$

$\begin{array}{c}1.49\times {10}^{-18}\text{ kg}\cdot \text{m/s}=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\left(1.67\times {10}^{-27}\text{ kg}\right)v\\ 1-\frac{v^2}{c^2}=\frac{{\left(1.67\times {10}^{-27}\text{ kg}\right)}^2{v}^2}{{\left(1.49\times {10}^{-18}\text{ kg}\cdot \text{m/s}\right)}^2}\\ 1-\frac{v^2}{c^2}=1.25\times {10}^{-18}{v}^2\\ 1=\frac{v^2}{c^2}+1.25\times {10}^{-18}{v}^2\end{array}$

Solve for $v$

$\begin{array}{c}\frac{v^2}{c^2}+\left(1.25\times {10}^{-18}{v}^2\times \frac{{\left(2.99\times {10}^8\right)}^2}{c^2}\right)=1\\ \frac{v^2}{c^2}+\frac{0.1117v^2}{c^2}=1\\ v^2=\frac{c^2}{1.1117}\\ v=0.948c\end{array}$

Conclusion:

The speed is $0.948c$ and the momentum of a proton is $1.49\times {10}^{-18}\text{ kg}\cdot \text{m/s}$.