A photosensitive surface has work function $\phi$. If photon of energy $3 \phi$ falls on this surface, the…

The relationship between incident photon energy $E$, work function $\phi$, and maximum photoelectron kinetic energy $K_{\text{max}}$ is governed by Einstein's photoelectric equation: $E = \phi + K_{\text{max}}$, where $K_{\text{max}} = \frac{1}{2} m v_{\text{max}}^2$.

For the first case, $E_1 = 3\phi$ and $v_1 = 4 \times 10^6 \text{ m/s}$ yields:

$3\phi = \phi + \frac{1}{2} m v_1^2 \Rightarrow 2\phi = \frac{1}{2} m v_1^2$

For the second case with $E_2 = 7\phi$:

$7\phi = \phi + \frac{1}{2} m v_2^2 \Rightarrow 6\phi = \frac{1}{2} m v_2^2$

Dividing these results eliminates the material-dependent constants:

$\frac{6\phi}{2\phi} = \frac{\frac{1}{2} m v_2^2}{\frac{1}{2} m v_1^2} \Rightarrow 3 = \frac{v_2^2}{v_1^2} \Rightarrow v_2 = \sqrt{3} v_1$

Substituting $v_1 = 4 \times 10^6 \text{ m/s}$ gives $v_2 = 4\sqrt{3} \times 10^6 \text{ m/s}$, corresponding to option A.