The distance between a light source and photoelectric cell is d. If the distance is decreased to `(d)/(2)` then

Options

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 electron per second will be doubled

Answer

A, C

Explanation

The intensity of light incident on the photoelectric cell,
`I prop (1)/(d^(2))`,
when the distance between the light source and photoelectric cell is d.
Again, number of photoelectrons emitted, `n prop I`
`:. n prop (1)/(d^(2)) or, (n_(2))/(n_(1)) = ((d_(1))/(d_(2)))^(2)`
`or, n_(2) = n_(1) xx ((d_(1))/(d_(2)))^(2) = n_(1) ((d)/((d)/(2)))^(2) = 4n_(1)`
Maximum kinetic energy of photoelectrons,
`E_("max") = eV_(0) = hf - W_(0)`
As the frequency (f) of incident light remains same, so the maximum kinetic energy `(E_("max"))` and stopping potential `(V_(0))` will remain unchanged.