Question

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

Answer 1

Potential of the collector, V=100 V
Magnetic field experienced by the electron, $B=2.83\times {10}^{-4}T$
Radius of the circular orbit, $R=12\text{cm}=12.0\times {10}^{-2}m$
From work-kinetic energy theorem
$\frac{1}{2}m{v}^2=eV$
$v^2=\frac{2eV}{m}$
$F=q(\overrightarrow{v}\times \overrightarrow{B})=qvB\sin \theta$
The angle between v and B is ${90}^{\circ }$.
$F=\text{qvB}$
As electron will undergo circular motion and magnetic force will provide essential centripetal force.
$\frac{m{v}^2}{R}=qvB$
$R=\frac{mv}{qB}=\frac{mv}{eB}$
$\frac{e}{m}=\frac{v}{BR}$
${\left(\frac{e}{m}\right)}^2=\frac{v^2}{B^2{R}^2}$
Putting value of $v^2{\left(\frac{e}{m}\right)}^2=\frac{\frac{2eV}{m}}{B^2{R}^2}$
$\frac{e}{m}=\frac{2V}{B^2{R}^2}$
Putting values, we get
$\frac{e}{m}=\frac{2\times 100}{(2.83\times {10}^{-4}\times 0.12{)}^2}$
$\frac{e}{m}=1.73\times {10}^{11}\text{C/kg}$
Therefore, the specific charge ration $\frac{e}{m}$ is $1.73\times {10}^{11}\text{C/kg}$
Final Answer: $1.73\times {10}^{11}\text{C/kg}$