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 becuase 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

Here, `V=100V, B=2.83xx10^(-4)T, r=12.0cm=12.0xx10^(-2)m`
when electrons are accelerated through V volt, the gain in K.E. of the electron is given by
`1/2mv^(2)=eV or v^(2)=(2eV)/m........(i)`
Since the electron moves in circular orbit under magnetic field, therefore, force on the electron due to magnetic field provides the centripetal force to the the electron.
`:. evB=mv^(2)//r ot eB=mv//r or v^(2)e^(2)B^(2)r^(2)//m^(2)..........(ii)`
From equation (i) and (ii), we get
`(2eV)/m=(e^(2)B^(2)r^(2))/(m^(2)) or e/m=(2V)/(r^(2)B^(2)) =(2xx100)/((12xx10^(-2))^(2)xx(2.83xx10^(-4))^(2))=1.73xx10^(11)Ckg^(-1)`