Question

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

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

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