Question

The distance between a light source and photoelectric cell is $d$. If the distance is decreased to $d/2$ 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 1

Answer: (A), (C)

The emission of electron per second will be four times
Stopping potential will remain same

$I\propto \frac{1}{r^2}$ and $I\propto N$(number of photons per second)

$⟹N\propto \frac{1}{r^2}$

Therefore, number of ejected electrons becomes $4$ times.

Also,

$K{E}_{max}=h\nu -\varphi$

Since $\nu$ remains unchanged, therefore, $K{E}_{max}$ as well as stopping potential remains unchanged.

$K{E}_{max}=e{V}_s$