The radiation corresponding to $3 \to 2$ transition of a hydrogen atom falls on a gold surface to generate photoelectrons. These electrons are passed through a magnetic field of $5\times10^{-4}$ T. Assume that the radius of the largest circular path followed by these electrons is 7 mm, the work function of the metal is:
(Mass of electron $= 9.1\times10^{-31}$ kg)
Answer: (D) 0.82 eV
Photon energy: $13.6\left(\frac14 - \frac19\right) = 1.89$ eV.
Fastest electrons: $r = \dfrac{mv}{eB} \Rightarrow v = \dfrac{eBr}{m} = \dfrac{1.6\times10^{-19}\times5\times10^{-4}\times7\times10^{-3}}{9.1\times10^{-31}} \approx 6.15\times10^5$ m/s
$$K_{max} = \frac12mv^2 = \frac12\times9.1\times10^{-31}\times(6.15\times10^5)^2 \approx 1.72\times10^{-19}\ \text{J} \approx 1.08\ \text{eV}$$
$\phi = 1.89 - 1.08 \approx 0.82$ eV.
Solution by Sreeraj P, M.Sc Physics