Photoelectrons are emitted from a metal surface when illuminated with UV light of wavelength 330 nm. The minimum amount of energy required to emit…

ii)i) Energy of incident radiation

$$\begin{aligned} & E=h v=h \frac{c}{\lambda} & =\frac{$\displaystyle 6.63$ \times $\displaystyle 10$^{-$\displaystyle 34$} \times $\displaystyle 3$ \times $\displaystyle 10$^{$\displaystyle 8$}}{$\displaystyle 330$ \times $\displaystyle 10$^{-$\displaystyle 9$}} & =$\displaystyle 6.027$ \times $\displaystyle 10$^{-$\displaystyle 19$} \mathrm{~J} \end{aligned}

$$ii) Kinetic energy of photoelectron

$$\begin{aligned} \text { K.E. } & =E-\phi_{$\displaystyle 0$} & =\left($\displaystyle 6$ \cdot $\displaystyle 027$ \times $\displaystyle 10$^{-$\displaystyle 19$}-$\displaystyle 3$ \cdot $\displaystyle 5$ \times $\displaystyle 10$^{-$\displaystyle 19$}\right) \mathrm{J} & =$\displaystyle 2$ \cdot $\displaystyle 527$ \times $\displaystyle 10$^{-$\displaystyle 19$} \mathrm{~J} \end{aligned}