Question

When monochromatic light from a bulb falls on a photosensitive surface, the number of photoelectrons emitted per second is $n$ and their maximum kinetic energy is $K_{max}$. If the distance of the lamp from the surface is halved, then

• A. $n$ is doubled
• B. $n$ becomes 4 times
• C. $K_{max}$ is doubled
• D. $K_{max}$ remains unchanged

Answer 1

Answer: (B), (D)

The correct options are
B $n$ becomes 4 times
D $K_{max}$ remains unchanged

K.E max doesn't depend on intensity but on frequency that remains unchanged and n becomes 4 times as intensity becomes due to its inverse square relation with distance.

So, the answers are option (B) & (D).