Photoelectric equation

For photoelectric emission, $K_{\max}=h\nu-\phi$ 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 $eV_0=K_{\max}$

Check Statement I

Statement I: Stopping potential in photoelectric effect does not depend on the power of the light source.

• Power/intensity of light mainly affects the number of photoelectrons emitted.
• It does not affect the energy of each photon.
• Therefore, it does not affect $K_{\max}$, and hence does not affect stopping potential $V_0$.

So, Statement I is correct.

Check Statement II

Statement II: For a given metal, the maximum kinetic energy of the photoelectron depends on the wavelength of the incident light.

Since $K_{\max}=h\nu-\phi$ and $\nu=\frac{c}{\lambda}$ therefore, $K_{\max}=\frac{hc}{\lambda}-\phi$

For a given metal, $\phi$ is constant, so $K_{\max}$ depends on $\lambda$.

Hence, Statement II is correct.

Choose the correct option

Since both statements are correct, the correct option is: $\boxed{\text{C}}$