For a photosensitive surface, work function is $3.3 × 10^{-19} J$. Taking Planck's constant to be $6.6 × 10^{-34} Js$, the threshold frequency would be

$5 × 10^{14} Hz$

The correct answer is Option (3) → $5 × 10^{14} Hz$

Given:

Work function, $\phi = 3.3 \times 10^{-19}\ \text{J}$

Planck's constant, $h = 6.6 \times 10^{-34}\ \text{Js}$

Threshold frequency is given by:

$f_{\text{th}} = \frac{\phi}{h}$

Substitute values:

$f_{\text{th}} = \frac{3.3 \times 10^{-19}}{6.6 \times 10^{-34}}$

$f_{\text{th}} = 0.5 \times 10^{15}\ \text{Hz} = 5 \times 10^{14}\ \text{Hz}$

∴ Threshold frequency = $5 \times 10^{14}\ \text{Hz}$