Use Einstein’s photoelectric equation

$K_{\max}=h\nu-\phi$

where:

• $K_{\max}$ = maximum kinetic energy of emitted photoelectrons
• $h\nu$ = energy of incident photon
• $\phi$ = work function of the metal

Also, stopping potential $V_0$ is related by

$eV_0=K_{\max}=h\nu-\phi$

Check statement A: The stopping potential depends only on the work function of the metal.

From $V_0=\frac{h\nu-\phi}{e}$ stopping potential depends on both frequency $\nu$ of incident light and work function $\phi$.

So, A is false.

Check statement B: The saturation current increases as the intensity of incident light increases.

Saturation current is proportional to the number of photoelectrons emitted per second. Higher intensity means more photons per second (for frequency above threshold), hence more emitted electrons.

So, B is true.

Check statement C: The maximum kinetic energy of a photoelectron depends on the intensity of the incident light.

From $K_{\max}=h\nu-\phi$ maximum kinetic energy depends on frequency, not on intensity.

So, C is false.

Check statement D: Photoelectric effect can be explained using wave theory of light.

Classical wave theory fails to explain important observations of the photoelectric effect, such as threshold frequency and dependence of $K_{\max}$ on frequency.

So, D is false.

Conclusion

Only statement B is correct.

Therefore, the correct option is:

$\boxed{\text{D: B only}}$

Comparison with stored answer

Stored correct answer = D

Derived answer = D

Hence, they agree.