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. Note: here I have explained only the sign of the energy, not its exact value. To avoid cluttering the derivation, some approximations have been made:

• $m_p \gg m_e$, so instead of the rotation of two particles about the CM, the rotation of the electron around the proton is considered;
• the orbit is assumed to be circular, not elliptical;
• no attention has been paid to Bohr's hypothesis about the quantisation of angular momentum.

If these approximations are removed, it can be shown that approximately $E_b \in [-13.6 ; 0)\ \text{eV}$.

But 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.