Q: 11.22  An electron gun with its collector at a potential of  $\dpi{100} 100\hspace{1mm}V$ fires out electrons in a spherical  bulb containing hydrogen gas at low pressure ($\dpi{100} \sim 10^-^2\hspace{1mm}mm$ of Hg). A magnetic field of  $\dpi{100} 2.83\times 10^4\hspace{1mm}T$  curves the path of the electrons in a circular orbit of radius $\dpi{100} 12.0\hspace{1mm}cm$ (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  $\dpi{100} e/m$  from the data.

S Sayak

The kinetic energy of an electron after being accelerated through a potential difference of V volts is eV where e is the electronic charge.

The speed of the electron will become

$v=\sqrt{\frac{2eV}{m_{e}}}$

Since the magnetic field curves, the path of the electron in circular orbit the electron's velocity must be perpendicular to the magnetic field.

The force due to the magnetic field is therefore Fb=evB

This magnetic force acts as a centripetal force. Therefore

$\\\frac{m_{e}v^{2}}{r}=evB\\ \\\frac{m_{e}v}{r}=eB\\ \frac{m_{e}}{r}\times \sqrt{\frac{2eV}{m_{e}}}=eB\\ \sqrt{\frac{e}{m_{e}}}=\frac{\sqrt{2V}}{Br}\\ \frac{e}{m_{e}}=\frac{2V}{r^{2}B^{2}}\\ \frac{e}{m_{e}}=\frac{2\times 100}{(2.83\times 10^{-4})^{2}\times (0.12)^{2}}\\ \frac{e}{m_{e}}=1.73\times 10^{11}C\ kg^{-1}$

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