The variation of stopping potential ( V _ 0 ) as a function of the...

The relation between the stopping potential and the frequency of the incident light is given by

$e{V}_{0} = hν - ϕ . . . \left(1\right)$

When ${V}_{0} = 0$,

$ϕ = hν . . . \left(2\right)$

Substitute the value of the frequency corresponding to zero stopping potential into equation (2) to calculate the required work function of the material.

$\begin{cases}ϕ & = & \dfrac{6 . 626 \times {10}^{- 34}J \cdot s \times 5 \times {10}^{14}Hz}{1 . 6 \times {10}^{- 19}J / eV} \\ & = & \dfrac{6 . 626 \times 5}{16} \approx 2 . 07eV\end{cases}$