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
Answer & explanation
Correct answer: option 3
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 = $5 \times 10^{14}\ \text{Hz}$