The given maximum velocity is v and frequency is n. We know that the kinetic energy is given by

$$KE = {1 \over 2}m{v^2} = hn - \phi $$

where h is Planck's constant and $$\phi$$ is the work function. Therefore, the kinetic energy of the incident light is

$${E_1} = hn - \phi $$ ..... (1)

When the frequency of the incident light is increased to 3n, then the kinetic energy is given by

$${1 \over 2}mv_1^2 = 3hn - \phi $$

$$ \Rightarrow {E_2} = 3hn - \phi $$ ..... (2)

Substituting $$hn = {E_1} + \phi $$ [from Eq. (1)] in Eq. (2), we get

$${E_2} = 3({E_1} + \phi ) - \phi $$

$$ \Rightarrow {E_2} = 3{E_1} + 2\phi $$

$$ \Rightarrow {1 \over 2}mv_1^2 = 3 \times {1 \over 2}m{v^2} + 2\phi $$

$$ \Rightarrow v_1^2 = 3{v^2} + 2\phi \times {2 \over m}$$

$$ \Rightarrow v_1^2 = 3{v^2} + {{4\phi } \over m}$$

Thus, the velocity is more than $$\sqrt 3 v$$.