1. Use Einstein’s photoelectric equation

For photoelectric emission,

$h\nu = h\nu_0 + eV_s$

where:

• $\nu$ = frequency of incident light
• $\nu_0$ = threshold frequency
• $V_s$ = stopping potential

So,

$\nu_0 = \nu - \frac{eV_s}{h}$

2. Find the frequency of incident light

Given wavelength,

$\lambda = 6630\,\text{\AA} = 6630 \times 10^{-10}\,\text{m} = 6.63 \times 10^{-7}\,\text{m}$

Now,

$\nu = \frac{c}{\lambda} = \frac{3 \times 10^8}{6.63 \times 10^{-7}}$

$\nu \approx 4.52 \times 10^{14}\,\text{s}^{-1}$

3. Compute $\dfrac{eV_s}{h}$

Using $e = 1.6 \times 10^{-19}\,\text{C}$, $V_s = 0.42\,\text{V}$, $h = 6.63 \times 10^{-34}\,\text{Js}$,

$\frac{eV_s}{h} = \frac{1.6 \times 10^{-19} \times 0.42}{6.63 \times 10^{-34}}$

$= \frac{0.672 \times 10^{-19}}{6.63 \times 10^{-34}}$

$\approx 1.01 \times 10^{14}\,\text{s}^{-1}$

4. Find threshold frequency

$\nu_0 = 4.52 \times 10^{14} - 1.01 \times 10^{14}$

$\nu_0 \approx 3.51 \times 10^{14}\,\text{s}^{-1}$

Given that threshold frequency is written as

$\nu_0 = x \times 10^{13}\,\text{s}^{-1}$

So,

$x = \frac{3.51 \times 10^{14}}{10^{13}} = 35.1$

Nearest integer:

$x = 35$

5. Comparison with stored answer

Stored correct answer = $35$

This matches the derived answer.