How electron recieves energy from electromagnetic wave?

Question

I know that if I oscillate an electron, it will produce electromagnetic wave, the energy I use to oscillate the electron is radiated as electromagnetic wave. But if an electron is kept in front of an electromagnetic wave, how will it accept the energy in the wave?

Answer 1

In the context of classical electrodynamics, you have a free charge within a time-varying electromagnetic field. For a plane electromagnetic wave, we know that the magnitude of the $\mathbf{B}$ field is $|\mathbf{E}|/c$, so I'll just consider the electric field. Then we have an electric field such as

$$\mathbf{E}=E_0\cos(\vec{k}\cdot\vec{r}-\omega t)\hat{u}.$$

You also have an electron of charge $-e$ and mass $m_e$, so through Newton's second law

$$\vec{F}=m\vec{a} \to -e\mathbf{E}=m_e\vec{a} , $$

therefore the acceleration is

$$\vec{a}= -\frac{e}{m_e}\mathbf{E}=-\frac{e}{m_e}E_0\cos(\vec{k}\cdot{r}-\omega t)\vec{u}. $$

The electron is accelerated by the electric field as the electromagnetic wave crosses it. This will cause the electron to produce its own radiation, eventually resulting in the scattering of electromagnetic waves. It is Thomson scattering - scattering off from a single electron. It does not change the frequency (or equivalently, energy of the photons), it redirects them. It is relevant when the energy of the photons is much smaller than the mass-energy of the electron, so $h\nu \ll m_ec^2\approx 511$ keV. This is why Thomson scattering is not relevant, for example, in a medical context (radiology and radiotherapy), since the photon energy in those cases is always in the keV or MeV range.

When the energy of the incoming photons increases, quantum mechanical effects start to be noticeable. For an unbound electron, the quantum mechanical scattering process is Compton scattering. It becomes increasingly relevant as the energy of the incoming photons increases, and it will change the frequency (energy) of the incoming radiation.