When monochromatic light falls on a photosensitive material, the number of photoelectrons emitted per second is n and their maximum kinetic energy is `K_(max)`. If the intensity of the incident light is doubled keeping the frequency same, then:

The frequency of incident light falling on a photosensitive metal plate is doubled, the KE of the emitted photoelectrons is

The frequency of incident light falling on a photosensitive metal plate is doubled, the KE of the emitted photoelectrons is

The photoelectros emitted by an illuminted surface have a maximum kinetic energy of 3.0 eV. If the intensity of the light is tripled, What is the maximum kinetic energy of photoelectrons now?

The frequency of incident light falling on a photosensitive metal plate is doubled, the K.E of the emitted photo-electrons is

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.

When monochromatic radiation of intensity I falls on a metal surface, the number of photoelectrons and their maximum kinetic are N and T respectively. If the intensity of radiation is 2 I, the number of emitted electrons and their maximum kinetic energy are respectively.

When monochromatic radiation of intensity I falls on a metal surface, the number of photoelectrons and their maximum kinetic are N and T respectively. If the intensity of radiation is 2 I, the number of emitted electrons and their maximum kinetic energy are respectively.

If the frequency of incident light falling on a photosensitive metal is doubled, the kinetic energy of the emitted photoelectron is