The work function of a metal is $h{v}_0$. Light of frequency $v$ falls on this metal. The photoelectric effect will take place only if

According to Einstein's photoelectric equation,  

$h\nu -W=K_{max}$, where $h=$Plank's constant, $\nu =$frequency of radiation, $W=$work function and $K_{max}=$maximum kinetic energy of photoelectron.

$\because K_{max}\ge 0$, as kinetic energy can't be negative.

$$\begin{gathered} \therefore h\nu -W\ge 0 \\ \Rightarrow h\nu \ge W \end{gathered}$$

 For the photoelectric effect to take place, the energy of radiation, $h\nu$ must be greater than the work function, $h{\nu }_0$ of the metal. That is,

$h\nu \ge h{\nu }_0\Rightarrow \nu \ge {\nu }_{0.}$