Question

Radiations of frequency ${10^{15}} Hz$ are incident on three photosensitive surfaces. A, B and C. Following observations are recorded,
(a) Surface A. No photoemission occurs.
(b) Surface B. Photoemission occurs but the photoelectrons have zero kinetic energy.
(c) Surface C. Photoemission occurs and photoelectrons have some KE.
Based on Einstein's photoelectric equation, explain the three observations.

Answer 1

In order to solve the above question we should know the concept of photoelectric effect and Einstein’s explanation of the photoelectric equation. Then we can find the solution for each case.

Complete step by step answer:
According to the Einstein's photoelectric equation which is given as ${K_{\max }} = h(\nu - {\nu _0})$, where h is the Planck’s constant, the following conclusions are describes the above observations:
(a) In the case of surface A, the value of threshold frequency ${v_0}$ is more than the value of radiation frequency $v$ (=${10^{15}}Hz$). Hence, no photoelectric emission takes place.
(b) In the case of surface B, the value of threshold frequency ${v_0}$ is equal to the value of radiation of frequency $v$ (=${10^{15}}Hz$). Therefore, the photoelectric emission which takes place but the value of kinetic energy of the photoelectron which is emitted is zero.
(c) In the case of surface C, the value of threshold frequency ${v_0}$ is lesser than the value of radiation frequency $v$ (=${10^{15}}Hz$). Thus results, the photoelectric emission which takes place and it carries some amount of kinetic energy of photoelectrons which emitted.

Note:
We know that light is made up of photons, Einstein gave the theory that when a light which consists of photons falls on the metal surface, the energy of the photon gets transferred to the electron. The frequency below this, no electron emission takes place. The value of h = Planck's constant is given by =$6.621 \times {10^{ - 34}}Js$.