The stopping potential V o (in volt) as a function of frequency ( v ) for a sodium emitter, is shown in the…

From Einstein's photoelectric equation, $h\nu =\varphi +e{V}_0$.

$\Rightarrow V_0=\frac{h\nu }{\mathrm{e}}-\frac{\varphi }{\mathrm{e}} ......\left(1\right)$

At $\nu =4\times {10}^{14} \mathrm{Hz}$, $V_0=0$.

Putting the value in equation $\left(1\right)$, we get

 $0=\frac{\mathrm{hv}}{\mathrm{e}}-\frac{\varphi }{\mathrm{e}}$

So, Work function $\left(\mathrm{\varphi }\right)=\frac{\left(\mathrm{h}\mathrm{\nu }\right)}{\mathrm{e}}\left(\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{V}\right)$
$=\frac{6.63\times {10}^{-34}\times \left(4\times {10}^{14}\right)}{1.6\times {10}^{-19}}$
$=1.66\mathrm{e}\mathrm{V}$