$7.4.16.$ 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:
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:
Which is what we wanted to show. Returning to the main problem, from the conservation of energy:
+
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.
#### Answer
$$
\boxed{\text{Impossible}}
$$
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### Statement
### Statement
$7.4.16.$ Is a radiationless capture of an electron by a free proton (formation of a hydrogen atom) possible?
$7.4.16.$ Is a radiationless capture of an electron by a free proton (formation of a hydrogen atom) possible?
### Solution
### 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:
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:
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:
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.
E_{2} = (M+m)c^2 + E_b, \quad E_b<0.
$$
$$
Let us prove that this energy is indeed negative.
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:
Consider the circular motion of the electron around the proton, and write Newton's second law:
Which is what we wanted to show. Returning to the main problem, from the conservation of energy:
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,
E_{k} = E_b < 0,
$$
$$
which is a contradiction. Therefore, radiationless capture is impossible. A photon must necessarily carry away part of energy.
which is a contradiction. Therefore, radiationless capture is impossible. A photon must necessarily carry away part of energy.
