The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630…

We know that, $K{E}_{max}=hf-\varphi =\frac{hc}{\lambda }-\varphi$

Or, $\frac{hc}{\lambda }-\varphi =e{V}_0$, where, $\varphi =h{\nu }_{\mathrm{th}}$ and $V_0$ is stopping potential.

So, threshold frequency is 

${\nu }_{\mathrm{th}}=\frac{c}{\lambda }-\frac{e{V}_0}{h}$

$=\frac{3\times {10}^8}{66330\times {10}^{-10}}-\frac{1.6\times {10}^{-19}\times 0.42}{6.63\times {10}^{-34}}$

$=\frac{3}{6.630}\times {10}^{15}-\frac{1.6\times 0.42}{6.63}\times {10}^{15}$

$={10}^{15}\left(\frac{3}{6.630}-\frac{1.6\times 0.42}{6.63}\right)=0.4524-0.1013$

${\nu }_{\mathrm{th}}=35.11\times {10}^{13}$