A metal surface is illuminated by a light of given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one-fourth of its original value, then the maximum KE of emitted photoelectrons will become.

A metal surface is illuminated by a light of given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one fourth of its original value, then the maximum kinetic energy of the emitted photoelectrons would become

A photosensitive plate is illuminated by green light and photoelectrons are emitted with maximum kinetic energy equal to 4 eV. If the intensity of the radiations is reduced to one fourth of its original value, then the maximum kinetic energy of the photo electrons will be :

When the frequency of light incident on a metallic plate is doubled the KE of the emitted photoelectron will be

Assertion: A photosensitive surface is illuminated by a monochromatic light of wavelength lambda and intensity I. The number of photoelectrons emitted per second is doubled when the intensity of light is doubled. However, the maximum speed of emitted electrons remains unchanged. Reason: The number of electrons emitted per second is directly proportional to the intensity of the incident light. The kinetic energy of the emitted photoelectrons is independent of the intensity of the light.

In photoelectric effect if the intensity of light is doubled then maximum kinetic energy of photoelectrons will become

When the intensity of radiation incident on photographic plate is increased keeping frequency constant, thenthe number and K.E. of photoelectrons emitted?

When the frequency fo light incident an a metallic plate is doubled , the 1KE of the emitted photoelectrons will be :