Question

A photon of energy $h\nu$ is absorbed by a free electron of a metal having work function $\phi < hv$.
Then:
A. the electron is sure to come out.
B. the electron is sure to come out with kinetic energy $h\nu-\phi$
C. either the electron does not come out or it comes out with a kinetic energy $h\nu-\phi$
D. it may come out with kinetic energy less than $h\nu-\phi$

Answer 1

Here, the energy of the photon and its relationship with the work function is given. From the laws of the photoelectric effect, this gives the emission of the electron due to electromagnetic radiation. We know that $E_{p}=\phi+E_{e}$, this can be used to find if the photoelectron or emitted, if emitted, then what is its energy.

Formula used:
$E_{p}=\phi+E_{e}$

Complete step by step answer:
We know that the photoelectric effect of the electron is the emission of the electron due to electromagnetic radiation. Here the continuous energy of the radiation is transferred to the electron, also called photoelectron. This photoelectron gets excited and leaves the metal, commonly called photoemission. This happens under the following conditions:
1. If the frequency of the incident light must be greater than the threshold frequency, then the intensity of the photoelectric current is proportional to the intensity of the light.
2. The photo-electronic current doesn’t depend on the intensity of the incident light.
3. There is no time lag between the falling of the incident light and the emission of the photoelectron.
Then the energy $E$ of the photoelectron is given by Planck as:
$E=h\nu=\dfrac{hc}{\lambda}$ Where, $h$ is the Planck’s constant, $\nu$ is the frequency of the incident light, $c$ is the speed of light and $\lambda$ is the wavelength of the incident light.
Also, the frequency of the incident light and the kinetic energy is related as, $E_{p}=\phi+E_{e}$, where $E_{p}$ is the energy of the photon, $\phi$ is a work function, and $E_{e}$ is the energy of the photoelectron.
Given that, $\phi < hv$. Then the energy of the photo-electron is less than or equal to $h\nu-\phi$.

Thus the answer is, D. it may come out with kinetic energy less than $h\nu-\phi$

Note:
Photoemission occurs only under the following conditions:
1. If the frequency of the incident light must be greater than the threshold frequency, then the intensity of the photoelectric current is proportional to the intensity of the light.
2. The photo-electronic current doesn’t depend on the intensity of the incident light.
3. There is no time lag between the falling of the incident light and the emission of the photoelectron.