Question

Answer the following giving reason:
(a) All the photoelectrons do not eject with the same kinetic energy when monochromatic light is incident on a metal surface.
(b) The saturation current in case (a) is different for different intensity.
(c) If one goes on increasing the wavelength of light incident on a metal sur face, keeping its intensity constant, emission of photoelectrons stops at a certain wavelength for this metal.

Hint:

The photoelectric effect depends on the frequency of light and the work function of the material. Intensity increases the number of photoelectrons, while wavelength controls whether electrons are emitted.

Answer 1

Photoelectric Effect – Reasoning Based Questions

(a) Why do all the photoelectrons not eject with the same kinetic energy when monochromatic light is incident on a metal surface?

Photoelectrons are emitted from various depths within the metal. Those originating from deeper layers expend energy to overcome internal resistance and collide with atoms before reaching the surface.

Consequently, the kinetic energy of photoelectrons is calculated as:
\( K.E. = hu - \phi - \text{energy lost inside the metal} \)

This energy loss results in varying kinetic energies among the emitted photoelectrons.

(b) Why is the saturation current different for different intensities of light?

Saturation current signifies the maximum photocurrent, achieved when all emitted electrons are collected. An increase in light intensity leads to a greater number of photons striking the metal per unit time, thus increasing the emission rate of photoelectrons.

As current is directly proportional to the number of charge carriers, the saturation current rises with increasing light intensity.

(c) Why does the emission of photoelectrons stop at a certain wavelength even when the intensity of light is kept constant?

The energy of a photon is defined by:
\( E = \frac{hc}{\lambda} \)

As the wavelength \( \lambda \) increases, the photon energy \( E \) diminishes. When \( \lambda \) reaches a point where \( E \) is less than the metal's work function (\( \phi \)), photons lack sufficient energy to eject electrons.

Therefore, photoemission ceases beyond a specific cut-off wavelength \( \lambda_0 \), determined by:
\( \lambda_0 = \frac{hc}{\phi} \)