In an experiment with photoelectric effect, the stopping potential,

In the photoelectric effect, the stopping potential $$V_0$$ is related to the maximum kinetic energy of emitted photoelectrons by:

$$ eV_0 = KE_{max} $$

Let us analyze each option:

Option 1: Stopping potential increases with intensity - Incorrect. Stopping potential depends on frequency, not intensity.

Option 2: Stopping potential decreases with intensity - Incorrect. Same reason as above.

Option 3: Stopping potential increases with wavelength - Incorrect. $$V_0 = \frac{h\nu - \phi}{e} = \frac{hc/\lambda - \phi}{e}$$. As wavelength increases, $$V_0$$ decreases.

Option 4: Stopping potential is $$\frac{1}{e}$$ times the maximum kinetic energy - Correct. From $$eV_0 = KE_{max}$$, we get $$V_0 = \frac{KE_{max}}{e}$$.

The correct answer is Option 4.