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 $$5 \times 10^{-4}$$ 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 $$= 9.1 \times 10^{-31}$$ kg)

The energy of the photon emitted in the $$3 \to 2$$ transition of hydrogen is $$E_{photon} = 13.6\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = 13.6 \times \frac{5}{36} \approx 1.89$$ eV.

These photons eject photoelectrons from gold. The maximum kinetic energy of photoelectrons is $$K_{max} = E_{photon} - \phi$$, where $$\phi$$ is the work function.

The electrons with maximum kinetic energy follow the largest circular path in the magnetic field. Using $$r = \frac{mv}{eB}$$, the maximum kinetic energy is $$K_{max} = \frac{p^2}{2m} = \frac{(eBr)^2}{2m}$$.

Substituting $$e = 1.6 \times 10^{-19}$$ C, $$B = 5 \times 10^{-4}$$ T, $$r = 7 \times 10^{-3}$$ m, $$m = 9.1 \times 10^{-31}$$ kg:

$$K_{max} = \frac{(1.6 \times 10^{-19} \times 5 \times 10^{-4} \times 7 \times 10^{-3})^2}{2 \times 9.1 \times 10^{-31}} = \frac{(5.6 \times 10^{-25})^2}{1.82 \times 10^{-30}} = \frac{3.136 \times 10^{-49}}{1.82 \times 10^{-30}} \approx 1.723 \times 10^{-19}$$ J $$\approx 1.077$$ eV.

The work function is $$\phi = E_{photon} - K_{max} = 1.89 - 1.077 \approx 0.81 \approx 0.82$$ eV.