The frequency and the intensity of a beam of light falling on the surface of a photoelectric material are increased by a factor of two. This will

Options

A. increase both, the maximum kinetic energy of the photo-electrons, as well as photoelectric saturation current by a factor of two.
B. increase the maximum kinetic energy of the photo-electrons by a factor greater than two and would increase the photoelectric saturation current by a factor of two.
C. increase the maximum kinetic energy of the photo-electrons by a factor greater than two and will have no effect on the magnitude of the photoelectric saturation current produced.
D. increase the maximum kinetic energy of the photo-electrons by a factor of two but will have no effect on the saturation photoelectric current.

Answer

C

Explanation

The maximum `K.E.` of ejected photoelectron is `(K.E)_(max)=hv-phi_(0)`
If the frequency of photon is doubled, maximum kinetic energy of photo electron becomes `(K.E.)'_(max)=2hv-phi_(0),((K.E.)'_(max))/((K.E.)_(max))=(2(hv-(phi_(0))/2))/(hv-phi_(0))gt2` Photo current `i prop "int ensity of beam"/(hv)`
If intensity and frequency both are doubled, the photocurrent remains same.