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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 Kmax. If the distance of the lamp from the surface is halved, then

A
n is doubled
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B
n becomes 4 times
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C
Kmax is doubled
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D
Kmax remains unchanged
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Solution

The correct options are
B n becomes 4 times
D Kmax 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).

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