Answer to Question #309922 in Physics for Almas

Question

Determination of charge/mass ratio of electron using narrow fine beam tube experiment objective

Answer 1

The change in kinetic energy of electron when it moves through potential difference $V$ is given by

$\frac{mv^2}{2}=eV$

Hence, the speed of electron when it enters in magnetic field

$v=\sqrt{2eV/m}$

The electron in magnetic field moves along a circular path of radius

$r=\frac{mv}{eB}=\frac{m\sqrt{2eV/m}}{eB}=\sqrt{(2mV)/(eB^2)}$

Finally

$\frac{e}{m}=\frac{2V}{r^2B^2}$