Motion of charged particles in an electric field > Interaction of charged particles > Problem 7.4.16

Statement

$7.4.16.$ Is a radiationless capture of an electron by a free proton (formation of a hydrogen atom) possible?

Solution

In the centre‑of‑mass frame, the total momentum of the particles before capture is zero, the interaction energy of the particles is zero, and they may have some kinetic energy:

$$
E_{1} = (M+m)c^2 + E_{k}, \quad E_{\text{kin}} \ge 0.
$$
After capture, according to the conservation of momentum, the total momentum must remain zero. If this happens without radiation, we end up with a single particle, and it cannot have kinetic energy. However, the hydrogen atom acquires a negative electrostatic binding energy:
$$
E_{2} = (M+m)c^2 + E_b, \quad E_b<0.
$$
Let us prove that this energy is indeed negative.
Consider the circular motion of the electron around the proton, and write Newton's second law:
$$
\frac{m_e v^2}{r} = \frac{e^2}{4\pi\varepsilon_0 r^2}
\quad\Longrightarrow\quad
m_e v^2 = \frac{e^2}{4\pi\varepsilon_0 r}.
$$

Kinetic energy:
$$
K = \frac{m_e v^2}{2} = \frac{1}{2}\frac{e^2}{4\pi\varepsilon_0 r}.
$$

Potential energy:
$$
U = -\frac{e^2}{4\pi\varepsilon_0 r}.
$$

Total energy:
$$
E_b = K + U = \frac{1}{2}\frac{e^2}{4\pi\varepsilon_0 r} - \frac{e^2}{4\pi\varepsilon_0 r}
= -\frac{1}{2}\frac{e^2}{4\pi\varepsilon_0 r} < 0.
$$

Which is what we wanted to show. Returning to the main problem, from the conservation of energy:
$$
E_{k} = E_b < 0,
$$
which is a contradiction. Therefore, radiationless capture is impossible. A photon must necessarily carry away part of energy.

Answer

$$
\boxed{\text{Impossible}}
$$