A metal surface is illuminated by radiation of wavelength $4500\ \text{Å}$. The ejected photoelectron enters a constant magnetic field of $2\ \text{mT}$ making an angle of $90^\circ$ with the magnetic field. If it starts revolving in a circular path of radius $2\ \text{mm}$, the work function of the metal is approximately
Answer: (A) $1.36\ \text{eV}$
Photon energy: $E = \dfrac{12400}{4500}\approx2.76\ \text{eV}$.
$r = \dfrac{mv}{eB}\Rightarrow v = \dfrac{eBr}{m} = \dfrac{1.6\times10^{-19}\times2\times10^{-3}\times2\times10^{-3}}{9.1\times10^{-31}}\approx7.0\times10^5\ \text{m s}^{-1}$.
$$K = \frac12mv^2 = \frac{(eBr)^2}{2m}\approx2.25\times10^{-19}\ \text{J}\approx1.41\ \text{eV}$$
$\phi = 2.76 - 1.41\approx1.36\ \text{eV}$.
Solution by Sreeraj P, M.Sc Physics