Correct Answer - D
As the electron bea is not deflected, then
`F_(m)=F_(e) implies Bev=Ee`
or `v=(E)/(B)` . . . (i)
As the electron moves from cathode to anode, its potential energy at the cathode appears as its kinetic energy at the anode. If V is the potential difference between the anode and cathode, then potential energy of the electron at cathode =eV. also kinetic energy of the electron at anode `=(1)/(2)mv^(2)`. according to law of conservation of energy , `(1)/(2)mv^(2)=eV`
or `vsqrt((2eV)/(m))` . . (ii)
From eqs. (i) and (ii), we have
`sqrt((2eV)/(m))=(E)/(B)implies (e)/(m)=(E^(2))/(2VB^(2))`.