Photoelectric equation and stopping potential

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 defined by $eV_0 = K_{\max}$ Hence, $V_0 = \frac{K_{\max}}{e} = \frac{h\nu - \phi}{e}$

Now check each option

A: It depends upon frequency of the incident light.

From $V_0 = \frac{h\nu - \phi}{e}$ clearly $V_0$ depends on frequency $\nu$.

So, A is true.

B: It is $1/e$ times the maximum kinetic energy of electrons emitted.

Since $eV_0 = K_{\max}$ therefore, $V_0 = \frac{K_{\max}}{e}$ So this statement is correct.

Hence, B is true.

C: It increases with increase in intensity of the incident light.

Stopping potential depends on the maximum kinetic energy of emitted electrons, which depends on frequency, not intensity.

Increasing intensity increases the number of emitted electrons (photoelectric current), but not their maximum kinetic energy.

So, C is not true.

D: It depends on the nature of emitter material.

Since work function $\phi$ depends on the material, $V_0 = \frac{h\nu - \phi}{e}$ so $V_0$ also depends on the material.

Thus, D is true.

Conclusion

The statement that is not true is: $\boxed{\text{C}}$