Question

6. When a monochromatic source of light is at a distance of 0.2 m from a photocell, the stopping potential
(cut off voltage) and the saturation current are found to be respectively I volt and 27 mA. If the same
source is placed at a distance 0.6 m from the cell, then
(A) the stopping potential will be 0.25 V and current will be 27 mA
(B) the stopping potential will be 1 volt and the current will be 3 mA
(C) the stopping potential will be 1 volt and the current will be 9 mA
(D) the stopping potential and the current will be same as before.​

Answer 1

Answer:

its B

Explanation:

lets begin

at first let's talk about stopping potential

K.E = hv - hv'

where v' is minimum frequency required to a photo electron

relation between kinetic energy of photo electron and stopping potential is

K.E = eV

where V is stopping potential

if we increase the distance between source and surface , frequency of the source is not going to change so K.E of photo electron remains same so does stopping potential .

but relation between intensity of incoming radiation and distance between source and surface is that intensity is inversely proportional to square of distance between source and surface.

so if I is intensity in first case and I' is intensity in second case

if we take ratio of both intensities then

I/I' = 0.6×0.6/0.2×0.2

I/I' = 9

so I' = I/9

now intensity is directly proportional to number of photons falling on surface of metal (photo current)

hence the photo current in second condition is 1/9 of the first part.

so the current becomes 1/9 × 27

= 3mA