Use Einstein’s photoelectric equation

For photoelectric emission, $K_{\max}=h\nu-\phi$ where $\phi$ is the work function of the metal.

Also, stopping potential $V_0$ is related to maximum kinetic energy by $eV_0=K_{\max}$

Check Statement-1

If ultraviolet light is replaced by X-rays, then the frequency $\nu$ of incident radiation increases significantly.

Since $K_{\max}=h\nu-\phi$ a larger $\nu$ gives a larger $K_{\max}$.

Then from $eV_0=K_{\max}$ the stopping potential $V_0$ also increases.

Hence, Statement-1 is true.

Check Statement-2

Statement-2 says that photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.

This is not correct as a general explanation of the spread in electron speeds.

Even if the incident light is monochromatic, emitted photoelectrons can still have kinetic energies ranging from nearly zero to $K_{\max}$. This happens because electrons originate from different depths and energy states inside the metal and may lose different amounts of energy before escaping.

So the spread in speeds is not due to a range of frequencies in the incident light.

Hence, Statement-2 is false.

Choose the correct option

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

Therefore, the correct option is $\boxed{\text{D}}$

Compare with stored answer

The stored correct answer is B, but our derivation gives D.

Therefore, I disagree with the stored answer. Statement-2 is false in standard photoelectric theory.