Question

light of wavelength lambda shines on a metal surface with intensity x and metal emits y electrons per second of average energy z what will happen y and z if x is doubled

Answer 1

Explanation:

The photoelectric current or the rate of emission of photoelectrons is directly proportional to the intensity of incident radiation. So, if we double the intensity of light x, the photoelectric current or electrons per second y is also doubled.

The negative potential at which photoelectric current becomes zero is called "stopping potential" or "cut-off potential" denoted by V

o

. Maximum kinetic energy E

k

is given by formula:

Ek=eVo

So, the maximum kinetic energy does not depend on the intensity of incident light. Therefore, z will remain same.

Answer 2

Answer:

If 'lambda' is halved, 'E' of photons will be doubled but value of K.E is less than double. No effect on number of ejected electroms.

If 'I' intensity is doubled number of ejected electrons become double. But no effect on 'K.E' of ejected electrons.