Photoelectrons are emitted with a maximum speed of $7\times {10}^5{\mathrm{ms}}^{-1}$ from a surface when light of frequency $8\times {10}^{14}\mathrm{Hz}$ is showered on it, the threshold frequency for this surface is (Consider $\mathrm{h}=6.6\times {10}^{-34}\mathrm{J}-\mathrm{s}$, mass of ${\mathrm{e}}^{-}=9.1\times {10}^{-31}\mathrm{kg}$)

$$\begin{gathered} {\mathrm{Speed}}_{\max }=7\times {10}^5\mathrm{m}/\mathrm{s} \\ \mathrm{Kinetic} \mathrm{energy} \mathrm{of} {\mathrm{e}}^{-}=\frac{1}{2}{\mathrm{mv}}^2=\frac{1}{2}\mathrm{m}\times {(7\times {10}^5)}^2 \\ =\frac{1}{2}\times 9.1\times {10}^{-31}\times 49\times {10}^{10}=9.1\times {10}^{-21}\times 49\times 0.5\mathrm{J} \end{gathered}$$
Given incident frequency $=8\times {10}^{14}\mathrm{Hz}$
$\therefore$From photoelectric equation ${\mathrm{k}}_{\max }=\mathrm{h\nu }-{\mathrm{h\nu }}_0$
$$\begin{gathered} \Rightarrow {\mathrm{h\nu }}_0=\mathrm{h\nu }-{\mathrm{k}}_{\max } \\ =6.6\times {10}^{-34}\times 8\times {10}^{14}-91\times 49\times 5\times {10}^{-23} \\ =[6.6\times 8\times {10}^{-20}-91\times 49\times 5\times {10}^{-23}] \\ ={10}^{-21}[66\times 8-91\times 49\times 5\times {10}^{-2}] \\ =[528-223]\times {10}^{-21}=6.6\times {10}^{-34}{\mathrm{\nu }}_0 \\ {\mathrm{\nu }}_0=\frac{305}{6.6}\times {10}^{-21}\times {10}^{34}=\frac{30.5}{6.6}\times {10}^{14}={\mathrm{\nu }}_0 \\ \Rightarrow 4.64\times {10}^{14}\mathrm{Hz} \end{gathered}$$