A photon of energy E ejects a photoelectron from a metal surface whose work function is $\phi_0$. If this electron enters into a uniform magnetic field B in a direction perpendicular to the field and describes a circular path of radius r, then the radius r is given by, (in the usual notation)
Answer & explanation
Correct answer: option 4
$\text{K.E. of electron} = E - \phi_0$
$\frac{1}{2}mv^2 = E - \phi_0$
$v = \sqrt{\frac{2(E-\phi_0)}{m}}$
$\text{In magnetic field: } r = \frac{mv}{eB}$
$r = \frac{m}{eB} \cdot \sqrt{\frac{2(E-\phi_0)}{m}}$
$= \frac{1}{eB}\sqrt{2m(E-\phi_0)}$
$r = \frac{\sqrt{2m(E-\phi_0)}}{eB}$