Show that the laws of momentum and energy conservation forbid the complete absorption of a photon by a free electron, (Note: This is not the Photoelectric effect in the photoelectric effect, the electron is not free: the metal participates in momentum and energy conservation).

The relativistic energy E for a particle with mass m is

$\begin{array}{l}\frac{\mathrm{hc}}{\lambda }\mathbf{+}{\mathbf{v}}_i{\mathbf{mc}}^2\mathbf{+}{\mathbf{v}}_f{\mathbf{mc}}^2\\ \mathbf{E}\mathbf{=}{\mathbf{\nu }}_a{\mathbf{mc}}^2\end{array}$

C: speed of light in vacuum,

${\mathrm{\nu }}_{\mathrm{a}}=\frac{1}{{\sqrt{1-\left(\frac{\mathrm{u}}{\mathrm{c}}\right)}}^2}$

Here velocity of the particle is u,

The relativistic momentum p of an object of mass m and velocity u is

$\mathrm{\rho }={\mathrm{\nu }}_{\mathrm{a}}\mathrm{mu}$

Here ${\nu }_a$the Lorentz factor

Energy E of a photon is constant $6.6\times {10}^{-34}\text{\hspace{0.17em}J}\cdot \text{s}$ and c being speed of the light in vacuum $=3.8\times {10}^8\frac{\text{m}}{\text{s}}$.

The momentum P of a photon of wavelength $\lambda$ is $\mathrm{p}=\frac{\mathrm{h}}{\mathrm{\lambda }}$.

If the electron is moving, it’ll start the interaction with some momentum and energy already.

Apply laws of conservation of momentum and energy equations would be:-

Momentum of the electron and photon in the initial and final stage is:-

${\mathrm{P}}_{\mathrm{p}}+{\mathrm{p}}_{\mathrm{ei}}+{\mathrm{p}}_{\mathrm{ef}}$ ….. (i)

Here ${\mathrm{P}}_{\mathrm{p}}$ is the momentum of photon in initial state.

${\mathrm{P}}_{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{\lambda }}$

Pei is the momentum of photons in initial state

${\mathrm{P}}_{\mathrm{ei}}={\mathrm{\nu }}_{\mathrm{i}}{\mathrm{m}}_{\mathrm{ui}}$

${\mathrm{P}}_{\mathrm{ef}}$ is the momentum of electron in final state

$\mathrm{Pef}={\mathrm{\nu }}_{\mathrm{f}}{\mathrm{mu}}_{\mathrm{f}}$

Substituting $P_p=\frac{h}{\lambda }$ ,$Pei={\nu }_{i}mui$and ${\mathrm{P}}_{\mathrm{ef}}={\mathrm{\nu }}_{\mathrm{f}}{\mathrm{m}}_{{\mathrm{u}}_{\mathrm{f}}}$

In (1) solve as:

$\frac{\mathrm{h}}{\mathrm{\lambda }}+{\mathrm{\nu }}_{\mathrm{i}}{\mathrm{mu}}_{\mathrm{i}}+{\mathrm{\nu }}_{\mathrm{f}}{\mathrm{mu}}_{\mathrm{f}}$ ….. (ii)

Kinetic energy of the electron and photon in the initial state is

${\mathrm{E}}_{\mathrm{p}}+{\mathrm{E}}_{\mathrm{ei}}={\mathrm{E}}_{\mathrm{ef}}$ …… (iii)

Here energy of the Electron is:$\mathrm{Ep}=\frac{\mathrm{hc}}{\mathrm{\lambda }}$

Energy of photon in the initial state

$\mathrm{Eei}={\mathrm{v}}_{\mathrm{i}}{\mathrm{mc}}^2$

Here, ${\mathrm{v}}_{\mathrm{i}}$is the frequency of photon in initial state

Energy of photon in final state is:$\mathrm{Eef}={\mathrm{v}}_{\mathrm{f}}{\mathrm{mc}}^2$

Here, role="math" localid="1660044467511" $v_f$is the frequency of photon in the final state. Substitute $\mathrm{Ep}=\frac{\mathrm{hc}}{\mathrm{\lambda }}$

$E_{ei}=v_{i}m{c}^2$and$\mathrm{Eef}={\mathrm{v}}_{\mathrm{f}}{\mathrm{mc}}^2$in (3) we get

$\frac{\mathrm{hc}}{\mathrm{\lambda }}+{\mathrm{v}}_{\mathrm{i}}{\mathrm{mc}}^2+{\mathrm{v}}_{\mathrm{f}}{\mathrm{mc}}^2$

Hence, that equation is only satisfied when both u’s are equal to c. But since no particle with mass can ever have a velocity equal to c that equation can never be satisfied. Therefore the initial assumptions that the electron can completely absorb a photon are false so the free electron completely absorb a photon.