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 xx 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.

Explanation

Here V=100 V ,
`B=2.83xx10^(-4)T`
`r=12cm=12xx10^(-2)m`
`implies`When electron is accelerated with V volt,kinetic energy acquired by electron, `(1)/(2)mv^(2)=eV`
`therefore v^(2)=(2eV)/(m)`……(1)
`implies`When electron moves in magnetic field centripetal force is provided by magnetic field,
`therefore (mv^(2))/(r)=Bev`
`therefore v=(Ber)/(m)`
`therefore v^(2)=(B^(2)e^(2)r^(2))/(m^(2))`...(2)
From equation (1) and (2) ,
`(2eV)/(m)=(B^(2)e^(2)r^(2))/(m^(2))`
`therefore (2V)/(B^(2)r^(2))=(e)/(m)`
`therefore (e)/(m)=(2xx100)/((2.83xx10^(-4))^(2)xx(0.12)^(2))`
`therefore (e)/(m)=1.73418xx10^(11)`
`therefore (e)/(m)~~1.73xx10^(11)Ckg^(-1)`