The frequency and the intensity of a beam of light falling on the surface of photo-electric material are 7, increased by a factor of two. This will

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increase the maximum kinetic energy of the photo-electrons, as well as photo-electric current by a factor of two

increase the maximum kinetic energy of the photo-electrons and would increase the photo-electric current by a factor of two

increase the maximum kinetic energy of the photo-electrons by a factor of two and will have no effect on the magnitude of the photoelectric current produced

not produce any effect on the kinetic energy of the emitted electrons but will increase the photo-electric current by a factor of two.

answer is B.

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Detailed Solution

12mv2max=hv−W when v is doubled (14/ remains the same), 12mv2max i.e., (K.E.) max is increased. The photo-electric current is directly proportional to the intensity of incident light.

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