A proton is accelerated from rest through a potential difference of $\mathbf{500}\mathbf{}\mathit{M}\mathit{V}\mathbf{.}\mathbf{}$

(a) What is its speed?

(b) Classical mechanics indicates that quadrupling the potential difference would double the speed. Were a classical analysis valid, what speed would result from a $\mathbf{2000}\mathbf{}\mathit{M}\mathit{V}$potential difference?

(c) What speed actually results?

To find the speed of a proton accelerated under a potential difference of $500MV$, we must equate relativistic kinetic energy to change in potential energy.

Relativistic kinetic energy is equal to change in potential energy as the charged particle is accelerated under a potential difference

localid="1659086760380" $$\begin{gathered} KE=∆U \\ KE=e∆V...................\left(1\right) \end{gathered}$$

Here, KE is the kinetic energy, $\Delta U$ is the change in potential energy, e is the charge, and $\Delta V$ is the potential difference.

Andas the electron is traveling at a very high speed considering relativistic Kinetic energy as below.

$\begin{array}{rcl}KE& =& Totalenergy-m_o{c}^2\\ & =& m{c}^2-m_o{c}^2\\ & =& \gamma m_o{c}^2-m_o{c}^2\\ & & \end{array}$

$KE=\left(\gamma -1\right)m_o{c}^2$ ….. (2)

Here, m₀ is the rest mass which is equal to the mass of the proton. Therefore, by comparing equation (1) and (2), you get

$\left(\gamma -1\right)m_p{c}^2=e\Delta V$ ….. (3)

Consider the known data as below.

The mass of proton, $m_p=1.67\times {10}^{-27}kg$

The speed of light, $c=3\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

The charge, $e=1.6\times {10}^{-19}C$

The potential difference, $\Delta V=500MV=5\times {10}^{8}V$

Substitute all known numerical values into equation (3), and you have

$$\begin{gathered} (\gamma -1)(1.67\times {10}^{-27}kg){(3\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.)}^2=(1.6\times {10}^{-19}C)(5\times {10}^{8}V) \\ (\gamma -1)=0.532 \\ \gamma =1.532 \end{gathered}$$

Then using formula of Lorentz factor

localid="1659087345441" $$\begin{gathered} \gamma =\frac{1}{\sqrt{1-{\displaystyle \raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}} \\ 1.532=\frac{1}{\sqrt{1-{\displaystyle \raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}} \\ 1-\raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.=\frac{1}{2.347} \\ \raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.=1-0.426 \\ {v}^2=0.574c^2 \\ v=0.758c \end{gathered}$$

Hence, at 500MV , the peed of a proton is 0.758c.

Here, we will equate classical kinetic energy to potential energy

$$\begin{gathered} \frac{1}{2}m{v}^2=eV \\ {v}^2=\frac{2eV}{m} \\ {v}^2=\frac{2\left(1.6\times {10}^{-19}C\right)\left(2\times {10}^9\right)}{1.67\times {10}^{-27}} \\ v\approx 2c \end{gathered}$$

Hence, at ,2000MV the classical speed is 2c .

According to the Einstein’s second postulate that, the speed of light is constant irrespective of the reference frame.

Therefore, classical result is invalid.

$\left(\gamma -1\right)m_p{c}^2=e\Delta V$

Here,

The mass of proton, $m_p=1.67\times {10}^{-27}kg$

The speed of light, $c=3\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

The charge, $e=1.6\times {10}^{-19}C$

The potential difference, $\Delta V=2000MV=2\times {10}^{9}V$

Substitute all known numerical values in the above equation.

$$\begin{gathered} \left(\gamma -1\right)\left(1.67\times {10}^{-27}kg\right){\left(3\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)}^2=\left(1.6\times {10}^{-19}C\right)\left(2\times {10}^{9}V\right) \\ \left(\gamma -1\right)\left(15.03\times {10}^{-11}\right)=\left(3.2\times {10}^{-10}\right) \\ \left(\gamma -1\right)=2.129 \\ \gamma =3.129 \end{gathered}$$

Then using formula of Lorentz factor as given by,

$$\begin{gathered} \gamma =\frac{1}{\sqrt{1-{\displaystyle \raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}} \\ 3.129=\frac{1}{\sqrt{1-{\displaystyle \raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}} \\ 1-\raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.=\frac{1}{9.791} \\ \raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.=1-0.102 \\ {v}^2=0.898c^2 \\ v=0.947c \end{gathered}$$

Hence, at 2000MV , the Speed of a proton is 0.947c.