Question

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?

Short answer

1. At $500MV$, speed of a proton is $0.7567c$.
2. At $2000MV$, classical speed is $2c$
3. The relativistic speed is $0.9476c$.

Step-by-step solution

Determine relation between kinetic energy and potential difference

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)

(a) Speed when potential difference is500MV.

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.

(b) Classical speed when potential difference is quadrupled that is V=2000MV.

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.

(c) Determine the relativistic speed to get the actual result:Rewrite equation (3) as below.

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

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.