(a) Frequency of photon $=\frac{\mathit{E}}{\mathit{h}}$
∴ Frequency $=\mathit{v}=\frac{6.5\times {10}^{-19}}{6.6\times {10}^{-34}}={10}^{15}\mathit{H}\mathrm{z}$
(b) work function of $\mathit{C}\mathit{s}={\mathit{\varphi }}_0=2.14\mathit{e}\mathit{V}$
Threshold frequency $={\mathit{v}}_0=\frac{{\mathit{\varphi }}_0}{\mathit{h}}$
$=\frac{2.14\times 1.6\times {10}^{-19}}{6.6\times {10}^{-34}}=0.5\times {10}^{15}\mathit{H}\mathrm{z}$
Since the energy of incident photon is more than the threshold frequency, emission of photoelectrons will be possible.
$\mathit{K}{\mathit{E}}_{\max }=$ Energy of incident photon $-$ Work function
$\therefore \mathit{K}{\mathit{E}}_{\max }=6.5\times {10}^{-19}-2.14\times 1.6\times {10}^{-19}$
$=3.1\times {10}^{-19}\mathit{J}$