Question

What percentage of the speed of light will an electron be moving after being accelerated through a potential difference of 1,844 Volts?

Answer 1

Answer:

The electron will be moving at 8.5% of the speed of light

Explanation:

The mass of an electron is:

$m=9.1\times10^{-31}kg$

The potential difference, V = 1844 Volts

The charge of an electron is:

$e=1.6\times10^{-19}$

The speed of the electron is calculated below

$\begin{gathered} v=\sqrt{\frac{2eV}{m}} \\ \\ v=\sqrt{\frac{2\times1.6\times10^{-19}\times1844}{9.1\times10^{-31}}} \\ \\ v=25464476.4\text{ m/s} \end{gathered}$

The speed of light is:

$c=8\times10^8\text{ m/s}$

The percentage of the speed of light that the electron will move is:

$\begin{gathered} \frac{25464476.4}{3\times10^8}\times100\% \\ \\ =8.5\% \end{gathered}$

The electron will be moving at 8.5% of the speed of light