When a light beam of frequency $\nu$ is incident on a metal surface, photoelectrons are emitted. If these
electrons describe a circle of radius r in a magnetic field of strength B, then the work function of the
metal is [EAMCET 2015]

When charged particle is moving on a circular path in a  uniform magnetic field B, then

$\frac{m{v}^2}{r}=qvB\Rightarrow \frac{mv}{r}=qB\Rightarrow v=\frac{qBr}{m}$

Thus, kinetic energy  $=\frac{1}{2}m{v}^2=\frac{1}{2}m\times {\left(\frac{qBr}{m}\right)}^2=\frac{1}{2m}(qBr{)}^2$

Now for photoelectric effect, Einstein's equation is

$h\mathrm{v}=W+\left(\mathrm{KE}\right)$

where, W = work function

$\Rightarrow hv-\mathrm{KE}=W\Rightarrow W=hv-\frac{(qBr{)}^2}{2m}$

Here, charged particle is a photoelectron denoted by e

$\therefore W=hv-\frac{(eBr{)}^2}{2m_e}$