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MHT CET2019Evening ShiftPhysicsDual Nature of MatterActual

A metal surface is illuminated by 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 the emitted photoelectrons would be

Options

  1. ATwice the original value
  2. BFour times the original value
  3. COne fourth of the original value
  4. DUnchanged

Correct answer

D. Unchanged

Step-by-step solution

The maximum kinetic energy of photoelectrons is given by K E m a x = h v - v 0 …. (i) Where h = Planck’s constant, v= frequency of radiation and v 0 = threshold frequency. It can be seen from Eq. (i), that the maximum KE of emitted photoelectrons is proportional to the frequency of radiation and is independent of the intensity of radiation, so it remains unchanged.

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