The threshold frequency for a certain photosensitive metal is $v_0$. When it is illuminated by light of frequency $v = 2v_0$, the maximum velocity of photoelectrons is $v_0$. What will be the maximum velocity of the photoelectrons when the same metal is illuminated by light of frequency $v = 5v_0$?

$2v_0$

As $v_0$ is the threshold frequency.

∴ Work function, $\phi_0=hv_0$

According to Einstein photoelectric equation

$\frac{1}{2}mv_{max}^2=hv-\phi_0$

Where hv is the incident energy, $\phi_0$ is the work function of the metal and $\frac{1}{2}mv_{max}^2$ is the maximum kinetic energy of the emitted photoelectron.

As per question

$\frac{1}{2}mv_0^2=h(2v_0)-hv_0=hv_0$  (i)

and $\frac{1}{2}mv'^2=5(5v_0)-hv_0=4hv_0$  (ii)

Divide (ii) by (i), we get

$\frac{v'^2}{v_0^2}=\frac{4}{1}$

$v'^2=4v_0^2$ or $v'=2v_0$