Energy of the incident photon on the metal surface is 3 W and then 5 W , where W is the work function of that metal. The ratio of velocities of emitted photoelectrons is

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Using Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by $K = E - W$ , where $E$ is the energy of the incident photon and $W$ is the work function.

For the first case, $E_{1} = 3 W$ :
$K_{1} = 3 W - W = 2 W$

For the second case, $E_{2} = 5 W$ :
$K_{2} = 5 W - W = 4 W$

Since kinetic energy $K = \frac{1}{2} m v^{2}$ , the velocity is proportional to the square root of kinetic energy, $v \propto K$ .

The ratio of velocities is:
$\frac{v _{1}}{v _{2}} = \frac{K _{1}}{K _{2}} = \frac{2 W}{4 W} = \frac{1}{2} = 1 : 2$

Answer: $1 : 2$

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