Question

Show that it is not possible for a photon to be completely absorbed by a free electron.

Answer 1

Hint:To prove this statement, we have to make use of the theory of conservation of energy since there is a collision involved between the photon and the free electron. The theory of conservation of energy states the total energy of the system before and after the collision is the same.

Complete step-by-step answer:
The photon is a bundle or packet of wave, that contains the energy proportional to the frequency of the wave contained in the wave.
Energy of one photon,
${E_p} \propto \upsilon $
$ \Rightarrow {E_p} = h\upsilon $
$ \Rightarrow {E_p} = \dfrac{{hc}}{\lambda }$
where h = Planck’s constant, c = speed of light wave and $\lambda $= wavelength of the light wave.
The energy possessed by a free electron is given by Einstein's theory of mass-energy equivalence.
Energy of electron,
${E_e} = m{c^2}$
where m = mass of electron and c = speed of light.
When the photon and electron collide and the photon is absorbed by the electron, this means that the total energy of the photon and electron before the collision will be equal to the energy of the electron after the collision, since the collision is assumed to be elastic and no energy is destroyed or lost in the process of collision.
If ${m_0}$ is the mass of the electron before the collision and ${m_1}$ is the mass of electron after the collision, we have –
${E_p} + {E_0} = {E_1}$
$\dfrac{{hc}}{\lambda } + {m_0}{c^2} = {m_1}{c^2}$
$ \Rightarrow \dfrac{h}{\lambda } + {m_0}c = {m_1}c$

Now, the terms ${m_0}c$ and ${m_1}c$ represent the momentum of the electron where the velocity v = c.
Since the electron is stationary initially, the velocity of the electron can be considered as zero.
$\therefore {m_0}c = 0$
This gives us –
$\dfrac{h}{\lambda } = {m_1}c$
The term ${m_1}c$ represents the momentum of the electron where the velocity v = c
So, if that were to happen, the speed of the electron must be equal to the speed of light which is impossible in Physics.
Therefore, it is not possible for a photon to be completely absorbed by a free electron.

Note:The expression $E = \dfrac{{hc}}{\lambda }$ originates from the fact that the velocity of any wave is the product of the frequency and wavelength of the wave.
As per Planck’s law, we have –

$E = h\upsilon $

where $\upsilon $ = frequency of light.
The speed of light represented by c, is equal to –

$c = \lambda \upsilon $
$ \Rightarrow \upsilon = \dfrac{c}{\lambda }$

On substituting for frequency, we get the expression –

$E = \dfrac{{hc}}{\lambda }$