# An electron is accelerated through some potential difference to a final kinetic energy of 1.95...

## Question:

An electron is accelerated through some potential difference to a final kinetic energy of 1.95 MeV. Using special relativity, determine the ratio of the electron\'s speed v to the speed of light c. What value would you obtain for this ratio if instead you used the classical expression for kinetic energy?

## Relativistic energy

If a particle of rest mass {eq}\displaystyle {m_0} {/eq} is in motion with a momentum {eq}\displaystyle {p} {/eq} and velocity {eq}\displaystyle {v} {/eq} then according to the special theory of relativity its energy is {eq}\displaystyle {E=\sqrt{(pc)^2+(m_0c^2)^2}} {/eq}. This may also be rewritten as {eq}\displaystyle {E=\gamma m_0c^2} {/eq} where {eq}\displaystyle {\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}} {/eq}. This shows that even if the particle is at rest it still has the rest mass energy {eq}\displaystyle {m_0c^2} {/eq}.

The kinetic energy of the electron is given to be {eq}\displaystyle {1.95\ MeV=1.95\times 1.6\times 10^{-13}\ J}=3.12\times 10^{-13}\ J {/eq}.

Using {eq}\displaystyle {E=\gamma m_0c^2} {/eq} and the fact that the rest energy of the electron {eq}\displaystyle {m_0c^2=0.511\ MeV} {/eq} we get,

{eq}\displaystyle {\gamma=\frac{1.95}{0.511}=3.82} {/eq}

Therefore,

{eq}\displaystyle { v=\sqrt{1-\frac{1}{\gamma^2}}\ c}={\sqrt{1-\frac{1}{3.82^2}}\ c=0.965 c} {/eq}.

{eq}\displaystyle {\color{blue}{\frac{v}{c}=0.965}} {/eq}.

If instead the classical expression for kinetic energy is used then,

{eq}\displaystyle { \frac{1}{2}m_0v^2=1.95\ MeV} {/eq}

That is,

{eq}\displaystyle { \frac{1}{2}m_0c^2\frac{v^2}{c^2}=1.95 MeV} {/eq}

That is,

{eq}\displaystyle { \frac{v^2}{c^2}=\frac{2\times 1.95}{0.511}=7.63} {/eq}

Or,

{eq}\displaystyle {v=2.76 c} {/eq}.

{eq}\displaystyle {\color{blue}{\frac{v}{c}=2.76}} {/eq}

Clearly, this faster than light speed would violate the well-established tenets of relativity. Thus the correct way to compute the velocity is by using the relativistic expression for the energy. 