The relation between threshold wavelength $\left({\lambda }_0\right)$ and work function $\left(\phi \right)$ of a metal is given by

$\mathrm{\phi } = \frac{\mathrm{hc}}{{\mathrm{\lambda }}_0}$

Putting the values, we get:

$$\begin{gathered} \Rightarrow 2 \mathrm{eV} = \frac{1240 \mathrm{eV} \mathrm{nm}}{{\mathrm{\lambda }}_0} (\because \mathrm{hc} = 1240 \mathrm{eV} \mathrm{nm}) \\ \Rightarrow {\mathrm{\lambda }}_0 = \frac{1240 }{2} \\ \Rightarrow {\mathrm{\lambda }}_0 = 620 \mathrm{nm} \end{gathered}$$