Radiation, with wavelength $6561$ Å falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of $3\times10^{-4}$ T. If the radius of the largest circular path followed by the electrons is $10$ mm, the work function of the metal is close to:
Answer: (C) $1.1$ eV
Photon energy: $E = \dfrac{12400}{6561} \approx 1.89$ eV.
The fastest electrons move on the largest circle, $r = \dfrac{p}{eB}$:
$$p = eBr = 1.6\times10^{-19}\times3\times10^{-4}\times10^{-2} = 4.8\times10^{-25}\ \text{kg m/s}$$
$$K_{\max} = \frac{p^2}{2m} = \frac{(4.8\times10^{-25})^2}{2\times9.1\times10^{-31}} = 1.27\times10^{-19}\ \text{J} \approx 0.79\ \text{eV}$$
$$\phi = 1.89 - 0.79 \approx 1.1\ \text{eV}$$
Solution by Sreeraj P, M.Sc Physics