Question

If the frequency of light in a photoelectric experiment is doubled, the stopping potential will
(a) be doubled
(b) be halved
(c) become more than double
(d) become less than double

Answer 1

(c) become more than double

According to Einstein's equation of photoelectric effect,
$$\begin{gathered} e{V}_0=hv-\phi \\ \Rightarrow V_0=\frac{hv-\phi }{e}...\left(1\right) \end{gathered}$$
Here, V₀ = stopping potential
v = frequency of light
$\phi$ = work function

Let the new frequency of light be 2ν and the corresponding stopping potential be V₀'.
Therefore,
$$\begin{gathered} e{V}_0\text{'}=2hv-\phi \\ {V}_0\text{'}=\frac{2hv-\phi }{e}...\left(2\right) \end{gathered}$$

Multiplying both sides of equation (1) by 2, we get:
$2V_0=\frac{2hv-2\phi }{e}...\left(3\right)$
Now if we compare (2) and (3), it can be observed that:
$$\begin{gathered} \frac{2hv-\phi }{e}>\frac{2hv-2\phi }{e} \\ \Rightarrow V_0\text{'}>2V_0 \end{gathered}$$
It is clear from the above equation that if the frequency of light in a photoelectric experiment is doubled, the stopping potential will be more than doubled.