Question

An electron gun with its collector at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (~${10}^{-2}$ mm of Hg). A magnetic field of $2.83\times {10}^{4}T$ curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the fine beam tube method.) Determine e/m from the data.

Answer 1

Given:

The potential of the electron gun is $100V$.

Magnetic field present in the region is $B=2.83\times {10}^{4}T$.

The radius of the circular orbit is $12cm=0.12m$

The energy of the electron gets converted into the kinetic energy as the electron moves through the field. Speed is given by:

$\frac{1}{2}m{v}^2=eV$

$v=\sqrt{\frac{2eV}{m_e}}$

The force due to the magnetic field is given by:

$F_b=evB$

Since the magnetic field makes the electron to rotate in the circular path. so, the magnetic force acts as a centripetal force. Therefore

$\frac{m_e{v}^2}{r}=evB$

$\frac{m_{e}v}{r}=eB$

$\sqrt{\frac{e}{m_e}}=\frac{\sqrt{2V}}{Br}$

$\frac{e}{m_e}=\frac{2\times 100}{(2.83\times {10}^{-4}{)}^2\times (0.12{)}^2}\frac{e}{m_e}=1.73\times {10}^{11}Ck{g}^{-1}$