Question

Photoelectric effect supports quantum nature of light because
(a) there is a minimum frequency below which no photoelectrons are emitted
(b) the maximum kinetic energy of photoelectrons depends only on the frequency of light and not on its intensity
(c) even when the metal surface is faintly illuminated the photoelectrons leave the surface immediately
(d) electric charge of the photoelectrons is quantised

Answer 1

• Photoelectric effect supports quantum nature of light because there is a minimum frequency below which no photoelectrons are emitted, Also the maximum kinetic energy of photoelectrons depends only on the frequency of light and not on its intensity and even when the metal surface is faintly illuminated the photoelectrons leave the surface immediately.

• This is because if light was only wave, a minimum frequency would not matter as the wave would provide low but continuous energy meaning there would be a lag in the ejection of the electrons. However, this is not seen in the experiments. Thus, photoelectric effect supports quantum nature of light.

The correct answers are options (a), (b) and (c).

Answer 2

Photoelectric effect supports quantum nature of light because there is a minimum frequency below which no photoelectrons are emitted, the maximum kinetic energy of photoelectrons depends only on the frequency of light and not on its intensity and even when the metal surface is faintly illuminated the photoelectrons leave the surface immediately.

Explanation:

• Based on the nature of light, the photoelectric effect can be described. There is no uniform spreading of energy in light as per the quantum nature of light that is present in the quanta or packets called as photons.
• A photon’s energy, $E=h v$, where v is the light’s frequency and h is constant of Planck.
• Above the threshold frequency, the photon’s energy is enough for electron emission from the metal surface.
• There will be no emission of photoelectron below as the photon’s energy is low. Hence, option (a) is correct.
• Now, the electron’s kinetic energy, $K=h v 0-\varphi$.
• So, the photoelectron’s kinetic energy is dependent only on the light’s frequency. Photon’s kinetic energy can be increased by increasing the photon’s energy or light’s frequency, which is according to $E=h v$. So, option (b) is also correct.
• When there is a faint illumination of the metal surface, there will be an emission of photoelectrons. This shows that there will be a less interaction of photons with the electrons.
• But energy is absorbed by few electrons from the incident photons and they move out. So, option (c) is correct.