In a photoelectric experiment light of wavelength 800   nm produces photoelectrons with the smallest de…

Kinetic energy of the electron having de Broglie wavelength $\lambda$ can be written as, $$\begin{gathered} {\lambda }_b=\frac{h}{p}=\frac{h}{\sqrt{2mK}} \\ \Rightarrow K=\frac{h^2}{2m{\left({\lambda }_b\right)}^2} \end{gathered}$$

For the given values, we can write:

$$\begin{gathered} \frac{hc}{\lambda }=K+\varphi \\ \Rightarrow \frac{hc}{800\times {10}^{-9}}=\frac{h^2}{2m\times {\left(1\times {10}^{-9}\right)}^2}+\varphi \\ \Rightarrow \varphi =\frac{hc}{800\times {10}^{-9}}-\frac{h^2}{2m\times {\left(1\times {10}^{-9}\right)}^2} \\ =\frac{\left(6.6\times {10}^{-34}\right)\times \left(3\times {10}^8\right)}{800\times {10}^{-9}}-\frac{{\left(6.6\times {10}^{-34}\right)}^2}{2\left(9.1\times {10}^{-31}\right)\times {\left(1\times {10}^{-9}\right)}^2} \\ =\left(2.47\times {10}^{-19}-2.39\times {10}^{-19}\right) \mathrm{J} \\ =0.05 \mathrm{eV} \end{gathered}$$

Note: Question & Option has been modified as information shared in paper was not correct.