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

A
increase the maximum kinetic energy of the photo-electrons, as well as photo-electric current by a factor of two
B
increase the maximum kinetic energy of the photo-electrons and would increase the photo-electric current by a factor of two
C
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
D
not produce any effect on the kinetic energy of the emitted electrons but will increase the photo-electric current by a factor of two.

Solution:

${(\frac{1}{2} {mv}^{2})}_{max} = hv - W when v is doubled (14 / remains the same), {(\frac{1}{2} {mv}^{2})}_{max} {i.e., (K.E.)}_{max} is increased.$

The photo-electric current is directly proportional to the intensity of incident light.

Solution:

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