Work Function of Metal from Photoelectron Speed Ratios

According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is given by: $K_{\max} = \frac{1}{2} m u^2 = E - \Phi$

where:

• $E = \frac{hc}{\lambda}$ is the energy of the incident photon,
• $\Phi$ is the work function of the metal,
• $m$ is the mass of an electron, and
• $u$ is the maximum speed of the emitted photoelectrons.

Given $hc = 1240\text{ eV nm}$, we can calculate the energy of the incident photons for both wavelengths: For $\lambda_1 = 248\text{ nm}$: $E_1 = \frac{1240\text{ eV nm}}{248\text{ nm}} = 5\text{ eV}$

For $\lambda_2 = 310\text{ nm}$: $E_2 = \frac{1240\text{ eV nm}}{310\text{ nm}} = 4\text{ eV}$

The corresponding maximum kinetic energies are: $K_{\max, 1} = \frac{1}{2} m u_1^2 = 5 - \Phi$ $K_{\max, 2} = \frac{1}{2} m u_2^2 = 4 - \Phi$

Given that the ratio of the maximum speeds is: $\frac{u_1}{u_2} = \frac{2}{1}$

The ratio of their maximum kinetic energies is: $\frac{K_{\max, 1}}{K_{\max, 2}} = \left(\frac{u_1}{u_2}\right)^2 = \left(\frac{2}{1}\right)^2 = 4$

Substituting the expressions for kinetic energy: $\frac{5 - \Phi}{4 - \Phi} = 4$

Solving for $\Phi$: $5 - \Phi = 4(4 - \Phi)$ $5 - \Phi = 16 - 4\Phi$ $3\Phi = 11$ $\Phi = \frac{11}{3}\text{ eV} \approx 3.67\text{ eV} \approx 3.7\text{ eV}$

Therefore, the work function of the metal is approximately $3.7\text{ eV}$.

Correct Option: (A)