Question

What percentage of the speed of light will an electron be moving after being accelerated through a potential difference of 531 Volts?

Answer 1

Given:

The potential difference applied to an electron is V = 531 V

The magnitude of the charge on an electron is

`q=1.6*10^(-19)\text{ C}`

The mass of the electron is

`m\text{ = 9.1}*10^(-31)\text{ kg}`

To find the percentage of the speed of light with the speed of an electron.

Step-by-step explanation:

The speed can be calculated by the formula

`\begin{gathered} (1)/(2)mv^2=qV \\ v=\sqrt{(2qV)/(m)} \end{gathered}`

On substituting the values, the speed will be

`\begin{gathered} v=\sqrt{(2*1.6*10^(-19)*531)/(9.1*10^(-31))} \\ =1.37*10^7\text{ m/s} \end{gathered}`

The percentage of the speed of light with the speed of the electron is

`\begin{gathered} (v)/(c)*100\%=(1.37*10^7)/(3*10^8)*100\% \\ =4.567\% \end{gathered}`

Thus, the percentage of the speed of light with the speed of the electron is 4.567%