In an experiment with photoelectric effect, the stopping potential,

In an experiment with photoelectric effect, the stopping potential,
  1. increases with increase in the intensity of the incident light
  2. decreases with increase in the intensity of the incident light
  3. increases with increase in the wavelength of the incident light
  4. is $\left(\frac{1}{\mathrm{e}}\right)$ times the maximum kinetic energy of the emitted photoelectrons

Solution

From Einstein photoelectric equation
$\frac{h c}{\lambda}=\phi+e V_S$
Maximum K.E. $=(\mathrm{K})_{\max }=e V s$
So, $V_S=\frac{(\mathrm{K})_{\max }}{e}$

Asked in: JEE Main 2025 (29 Jan Shift 2)

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