Given:
The potential difference applied to an electron is V = 531 V
The magnitude of the charge on an electron is
[tex]q=1.6\times10^{-19}\text{ C}[/tex]The mass of the electron is
[tex]m\text{ = 9.1}\times10^{-31}\text{ kg}[/tex]To find the percentage of the speed of light with the speed of an electron.
Explanation:
The speed can be calculated by the formula
[tex]\begin{gathered} \frac{1}{2}mv^2=qV \\ v=\sqrt{\frac{2qV}{m}} \end{gathered}[/tex]On substituting the values, the speed will be
[tex]\begin{gathered} v=\sqrt{\frac{2\times1.6\times10^{-19}\times531}{9.1\times10^{-31}}} \\ =1.37\times10^7\text{ m/s} \end{gathered}[/tex]The percentage of the speed of light with the speed of the electron is
[tex]\begin{gathered} \frac{v}{c}\times100\%=\frac{1.37\times10^7}{3\times10^8}\times100\% \\ =4.567\% \end{gathered}[/tex]Thus, the percentage of the speed of light with the speed of the electron is 4.567%
