Let $$K_1$$ and $$K_2$$ be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength $$\lambda_1$$ and $$\lambda_2$$, respectively are incident on a metallic surface. If $$\lambda_1 = 3\lambda_2$$ then:

Using Einstein's photoelectric equation, the maximum kinetic energy of photo-electrons is given by $$K = \frac{hc}{\lambda} - \phi$$, where $$\phi$$ is the work function of the metal. For wavelength $$\lambda_1$$ and $$\lambda_2$$ we have $$K_1 = \frac{hc}{\lambda_1} - \phi$$ and $$K_2 = \frac{hc}{\lambda_2} - \phi$$. Given that $$\lambda_1 = 3\lambda_2$$, substitution into the expression for $$K_1$$ yields $$K_1 = \frac{hc}{3\lambda_2} - \phi$$.

Now compute $$\frac{K_2}{3}$$: $$\frac{K_2}{3} = \frac{1}{3}\left(\frac{hc}{\lambda_2} - \phi\right) = \frac{hc}{3\lambda_2} - \frac{\phi}{3}$$. Comparing $$K_1$$ and $$\frac{K_2}{3}$$ gives $$K_1 - \frac{K_2}{3} = \left(\frac{hc}{3\lambda_2} - \phi\right) - \left(\frac{hc}{3\lambda_2} - \frac{\phi}{3}\right) = -\phi + \frac{\phi}{3} = -\frac{2\phi}{3}$$. Since $$\phi > 0$$, it follows that $$K_1 - \frac{K_2}{3} = -\frac{2\phi}{3} < 0$$, hence $$K_1 < \frac{K_2}{3}$$.

Answer: Option B.