Use Einstein’s photoelectric equation

For incident light of frequency $\nu$ on a metal of threshold frequency $\nu_0$, $K_{\max} = h\nu - h\nu_0 = h(\nu-\nu_0).$

Also, stopping potential $V_s$ is related by $eV_s = K_{\max} \quad \Rightarrow \quad V_s = \frac{K_{\max}}{e} = \frac{h}{e}(\nu-\nu_0).$

Check Statement-1

Initially, $K_{\max} = h(\nu-\nu_0).$

If the frequency is doubled, $\nu \to 2\nu$, then $K'_{\max} = h(2\nu-\nu_0).$

For Statement-1 to be true, we would need $K'_{\max} = 2K_{\max}.$

But $2K_{\max} = 2h(\nu-\nu_0)=h(2\nu-2\nu_0),$ whereas $K'_{\max}=h(2\nu-\nu_0).$

These are not equal in general.

In fact, $K'_{\max}-2K_{\max}=h\nu_0 \neq 0.$

So $K_{\max}$ does not double.

Since $V_s = \frac{K_{\max}}{e},$ stopping potential also does not double in general.

Therefore, Statement-1 is false.

Check Statement-2

From the equations, $K_{\max} = h\nu - h\nu_0,$ which is linear in $\nu$.

Also, $V_s = \frac{h}{e}\nu - \frac{h\nu_0}{e},$ which is also linear in $\nu$.

Therefore, Statement-2 is true.

Choose the correct option

• Statement-1: False
• Statement-2: True

Hence, the correct option is: $\boxed{\text{C}}$