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)

$\sqrt{\frac{2m(E-\phi_0)}{eB}}$

As the electron describes a circular path of radius r in the magnetic field, therefore,

$\frac{mv^2}{r}=evB;r=\frac{mv}{eB}=\frac{p}{eB}=\frac{\sqrt{2mK}}{eB}$  (As $K=\frac{p^2}{2m}$)

From Einstein’s photoelectric equation

$K=E-\phi_0$

$∴r=\sqrt{\frac{2m(E-\phi_0)}{eB}}$