WBJEE · Physics · Dual Nature of Matter
The distance between a light source and photoelectric cell is d. If the distance is decreased to \(\frac{d}{2},\) then
- A the emission of electron per second will be four times
- B maximum kinetic energy of photoelectrons will be four times
- C stopping potential will remain same
- D the emission of electrons per second will be doubled
Answer & Solution
Correct Answer
(C) stopping potential will remain same
Step-by-step Solution
Detailed explanation
The intensity of emitted electrons is \[ \begin{array}{l} I=\frac{1}{r^{2}} \\ I_{1}=\frac{1}{d^{2}} \\ I_{2}=\frac{1}{(d / 2)^{2}}=4 \cdot \frac{1}{d^{2}} \end{array} \] and I \(\propto\) Number of photons per second \(=N\) \(\therefore\) \[ N=\frac{1}{r^{2}} \] \(\therefore\)…
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