Question

In photoelectric effect if the intensity of light is doubled, then maximum kinetic energy of photoelectrons will become.
A. Double
B. Half
C. Four times
D. No change

Answer 1

Hint: The energy of the photon is given by the equation: \[E = h\upsilon \]. The intensity of light can be defined as the amount of photon energy per unit area. Thus the greater the intensity of light the more will be the number of photons. So as a result more will be the number of ejected electrons.

Formula used:
The maximum kinetic energy of photoelectrons is given as:
\[K{E_{\max }} = h\upsilon - {W_0}\]
Where \[h\] is the Plank constant, \[\upsilon \] is the frequency of incident light and \[{W_0}\] is the work function.

Complete step by step solution:
\[K{E_{\max }} = h\upsilon - {W_0}\]
Here \[h\upsilon \] is not dependent on the intensity of light. If we increase the intensity of light then the number of photons increases. Due to this, the emission of electrons increases.

So the maximum kinetic energy is dependent on energy not on intensity. If we change the frequency also the maximum kinetic energy changes. The kinetic energy of photoelectrons does not depend upon the intensity of light. Therefore in the photoelectric effect, if the intensity of light is doubled there will be no change in the maximum kinetic energy.

Hence option D is the correct answer.

Note: The photoelectric effect occurs when electrons are released from a metal surface when sufficient frequency light is impressed on it. When a photon collides with a metal surface, its energy is transferred to the electron. A photon's total energy is the sum of the energy used to expel an electron and the maximal kinetic energy of electrons.