An electron gun with its anode at a potential of 100V fires out electrons in a spherical bulb containing hydrogen gas at low pressure `(10^(-2)nm of Hg)`. A magnetic field of `2.83xx10^(-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 'fine beam tube' method). Determine `e//m` from the data.

Explanation

Given , potential , V = 100 V
Magnetic field , `B= 2.83xx10^(-4)T`
Radius , r = 12.0 cm `=12 xx10^(-2)` m
We know that when electrons are accelerated at potential difference V, gain in kinetic energy is given by
`eV=1/2mv^2" "...(i)`
Magnetic force on an electron moving under the influence of magnetic field provides the necessary centripetal force.
`:. Bev = (mv^2)/r " or "v = (eBr)/m" "...(ii)`
From (i) and (ii) ,
`e/m=(2V)/(r^2B^2)`
Putting the values
`e/m=(2xx100)/((12xx10^(-2))^2xx(2.83xx10^(-4))^2)`
`e/m=1.73xx10^(11)"C kg"^(-1)`