Question

When photons of energy $h\upsilon $ are incident on the surface of a photosensitive material of work function $h{{\upsilon }_{0}}$,then
A) $\text{The kinetic energy of all emitted electrons is }h{{\upsilon }_{0}}$.
B) $\text{The kinetic energy of all emitted electrons is }\,h\left( \upsilon -{{\upsilon }_{0}} \right)$.
C) $\text{The kinetic energy of all fastest electrons is }h\left( \upsilon -{{\upsilon }_{0}} \right)$.
D) $\text{The kinetic energy of all emitted electrons is }h\upsilon $.

Answer 1

Hint: Photoelectric theory explains how a photon helps in the ejection of an electron in metal having a certain work function. The work function is the least energy needed to emit an electron from the surface of the metal. A photon has an energy that is proportional to its frequency.

Complete step by step answer:
Suppose the electromagnetic radiation of light falling on a metal surface has a frequency $\upsilon $. So, the energy associated with the radiation is given by $h\upsilon $. Where, $h$ is the Planck’s constant.
The work function of a metal is defined as the minimum energy that is required to release an electron from its surface. So, we can compare that energy with the energy of a photon. So, suppose a photon of $h{{\upsilon }_{0}}$ is the minimum energy required by a photon in order to release an electron from the metal surface. We call this energy as the work function of the metal and the frequency ${{\upsilon }_{0}}$ is the threshold frequency.
So when a photon of frequency $\upsilon $ is falling on metal is greater than the threshold frequency, the electrons are emitted from the surface of the metal and these electrons will have kinetic energy that is equal to the difference between the energy of the photon and the work function of the metal.
$\text{Kinetic Energy (K}\text{.E)}=\text{Energy of the Photon}-\text{Work Function}$
$K.E=h\upsilon -h{{\upsilon }_{0}}$
So all the emitted electrons will have a kinetic energy in the range from zero to $h\left( \upsilon -{{\upsilon }_{0}} \right)$. This is because electrons which are lying inside the metal require more energy to come out of the metal compared to the electrons lying on the surface of the metal. So the Kinetic energy gained by these electrons inside the metal will be less than the kinetic energy gained by the electrons when they are out from the metal.
So, the answer to the question is option (C)- The kinetic energy of all fastest electrons is $h\left( \upsilon -{{\upsilon }_{0}} \right)$.

Note: In photo-electric effect the emission of an electron is instantaneous. As soon as the radiation above or equal to the work function falls on the metal, the electrons are emitted.
Electrons emitted by the photoelectric effect are called photoelectrons. The kinetic energy of an emitted electron is independent of the intensity of the radiation.
But intensity will increase the rate of emission of photoelectrons for a metal of given work function and radiation frequency.