Question

Through what potential difference must an electron be accelerated, starting from rest, to acquire a speed of $0.99c$?

Short answer

The potential difference required is$3.12\times {10}^6\mathrm{V}$.

Step-by-step solution

Given Information

We have given that, speed$v=0.99c$.

We have to find the potential difference.

Simplify

The mass of an electron is $m=9.10\times {10}^{-31}$, the kinetic energy of an electron travelling at a constant velocity$v=0.99c$is given by

role="math" localid="1649909533780" $$\begin{gathered} E_k=m{c}^2\left[\frac{1}{\sqrt{1-(v/c{)}^2}}-1\right] \\ =9.1\times {10}^{-31}\times {\left(3\times {10}^8\right)}^2\left[\frac{1}{\sqrt{1-(0.99{)}^2}}-1\right] \\ =4.99\times {10}^{-13}\mathrm{J} \end{gathered}$$

Now, if there is a potential difference $v$then we have

role="math" localid="1649909731842" $$\begin{gathered} vq=E_k \\ v=\frac{E_k}{q} \\ =\frac{4.99\times {10}^{-13}}{1.6\times {10}^{-19}} \\ =3.12\times {10}^{6}v \end{gathered}$$

So the potential difference required is $3.12\times {10}^6\mathrm{V}$.