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?

Even though monochromatic light (single frequency) has photons of the same energy, photoelectrons are emitted from different depths of the metal surface. Those emitted from deeper layers lose some energy in overcoming internal resistance and collisions with atoms before reaching the surface.

Therefore, the kinetic energy of photoelectrons is given by:
\( K.E. = h\nu - \phi - \text{energy lost inside the metal} \)

Hence, photoelectrons do not have identical kinetic energies.

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

Saturation current corresponds to the maximum photocurrent when all emitted electrons are collected. As the intensity of light increases, more photons are incident on the metal per unit time, resulting in the emission of more photoelectrons.

Since current is directly proportional to the number of electrons, saturation current increases with 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 given by:
\( E = \frac{hc}{\lambda} \)

As the wavelength \( \lambda \) increases, the energy \( E \) of each photon decreases. When \( \lambda \) becomes large enough that \( E \lt \phi \) (work function of the metal), photons no longer have enough energy to eject electrons.

Hence, photoemission stops beyond a certain cut-off wavelength \( \lambda_0 \), where:
\( \lambda_0 = \frac{hc}{\phi} \)