The stopping potential for photoelectrons emitted from a surface...

We know that, K{E}_\max = hf - ϕ = \dfrac{hc}{\lambda} - ϕ

Or, $\dfrac{hc}{\lambda} - ϕ = e{V}_{0}$, where, $ϕ = h{ν}_{th}$ and ${V}_{0}$ is stopping potential.

So, threshold frequency is

${ν}_{th} = \dfrac{c}{\lambda} - \dfrac{e{V}_{0}}{h}$

$= \dfrac{3 \times {10}^{8}}{66330 \times {10}^{- 10}} - \dfrac{1 . 6 \times {10}^{- 19} \times 0 . 42}{6 . 63 \times {10}^{- 34}}$

$= \dfrac{3}{6 . 630} \times {10}^{15} - \dfrac{1 . 6 \times 0 . 42}{6 . 63} \times {10}^{15}$

$= {10}^{15}\left(\dfrac{3}{6 . 630} - \dfrac{1 . 6 \times 0 . 42}{6 . 63}\right) = 0 . 4524 - 0 . 1013$

${ν}_{th} = 35 . 11 \times {10}^{13}$