Energy of the incident photons on the metal surface is initially 4 W and then 6 W where W is the work…

The kinetic energy of photoelectrons is given by $K_{\max} = \frac{1}{2}mv^{2}$ for incident energy $E$ exceeding the work function $W$, with $K_{\max} = E - W$.

For incident energy $E_1 = 4W$, the kinetic energy is $K_{\max1} = 4W - W = 3W$, so $\frac{1}{2}mv_1^{2} = 3W$.

For incident energy $E_2 = 6W$, the kinetic energy is $K_{\max2} = 6W - W = 5W$, so $\frac{1}{2}mv_2^{2} = 5W$.

Taking the ratio of the two kinetic energy equations yields $\frac{\frac{1}{2}mv_1^{2}}{\frac{1}{2}mv_2^{2}} = \frac{3W}{5W}$, which simplifies to $\frac{v_1^{2}}{v_2^{2}} = \frac{3}{5}$.

The velocity ratio is therefore $\frac{v_1}{v_2} = \sqrt{\frac{3}{5}} = \frac{\sqrt{3}}{\sqrt{5}}$.

Thus, the ratio of the velocities is $\sqrt{3}:\sqrt{5}$.