Edits to “Statement”, “Solution”, “Answer”

jzmicer edited
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### Statement
−$7.4.16.$ [Insert the problem statement]
+$7.4.16.$ Is a radiationless capture of an electron by a free proton (formation of a hydrogen atom) possible?
### Solution
−1
+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:
−#### Answer
+$$
+E_{1} = (M+m)c^2 + E_{k}, \quad E_{\text{kin}} \ge 0.
+$$
+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:
+$$
+\frac{m_e v^2}{r} = \frac{e^2}{4\pi\varepsilon_0 r^2}
+\quad\Longrightarrow\quad
+m_e v^2 = \frac{e^2}{4\pi\varepsilon_0 r}.
+$$
−[Insert a concise answer or boxed result]
+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
+$$
+\boxed{\text{Impossible}}
+$$