Photoelectric equation

In the photoelectric effect, the maximum kinetic energy of emitted electrons is $K_{\max} = h\nu - \phi$ where:

• $h\nu$ = energy of incident photon
• $\phi$ = work function of the metal

The stopping potential $V_0$ is related by $eV_0 = K_{\max} = h\nu - \phi$

Hence, for a given metal, $V_0 = \frac{h\nu - \phi}{e}$

So, the stopping potential depends on frequency of incident light, not on intensity.

Check Assertion A

Assertion A says: In photoelectric effect, on increasing the intensity of incident light the stopping potential increases.

This is false, because increasing intensity only increases the number of photons incident per second, not the energy per photon. Therefore, $K_{\max}$ and hence stopping potential do not increase.

Check Reason R

Reason R says: Increase in intensity of light increases the rate of photoelectrons emitted, provided the frequency of incident light is greater than threshold frequency.

This is true. If $\nu > \nu_0$, then increasing intensity means more photons fall per second, so more electrons are emitted per second, i.e. photoelectric current increases.

Conclusion

• Assertion A: False
• Reason R: True

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