Q.Consider aiming a beam of free electrons towards free protons. When they scatter, an electron and a proton cannot combine to produce a H-atom,

The key idea is that when a free electron and a free proton scatter, they can form a hydrogen atom only if the excess energy (the binding energy released) is carried away — otherwise energy conservation fails.

Reasoning:

1. A free electron and proton have total energy greater than the ground-state energy of hydrogen ($-13.6\ \text{eV}$). To bind into an atom, the system must lose this excess energy.
2. Without a third body to carry away the surplus, both energy and momentum cannot be simultaneously conserved in a two-body capture process. …

The key insight here is that when a free electron and a free proton come together to form a hydrogen atom, the total energy of the system decreases. The electron goes from having positive kinetic energy (in free space) to being in a bound state with negative total energy. That energy difference — the binding energy — has to go somewhere. It cannot simply vanish.

But there is a deeper constraint. In any scattering or capture process, both energy and momentum must be conserved. If the electron and proton simply combined without emitting anything, the final hydrogen atom would have to carry away all the initial momentum. That is possible in principle, but the problem is that the atom's kinetic energy would then be too large — it would exceed the binding energy released, making the capture impossible without a third particle to carry away the excess.

Let us walk through the reasoning step by step.

1. Energy conservation alone does not forbid capture. The electron and proton have some initial kinetic energy (say, $K_i$). When they bind into a hydrogen atom, the final energy is the atom's ground state energy ($-13.6\ \text{eV}$) plus its kinetic energy as a whole ($K_f$). Energy conservation gives

where $E_{\text{emitted}}$ is any energy carried away by radiation. If no radiation is emitted, then $K_f = K_i + 13.6\ \text{eV}$, which is perfectly allowed by energy conservation alone. So option (A) is false — energy conservation does not prevent capture.

2. Momentum conservation is the real obstacle.

   In the centre-of-mass frame, the total momentum before capture is zero. After capture, if no radiation is emitted, the hydrogen atom must also have zero momentum — so it is at rest. That means its kinetic energy $K_f = 0$. But then energy conservation would require $K_i = -13.6\ \text{eV}$, which is impossible since kinetic energy cannot be negative.

   In the lab frame, the situation is even clearer: the initial momenta of the electron and proton are generally not equal and opposite. If they combine into a single atom, the atom's momentum must equal the total initial momentum. That is fine — but the atom's kinetic energy is then $p^2/(2M)$, where $M$ is the atom's mass. This is tiny compared to the binding energy (since $M$ is large), but the binding energy is fixed at $13.6\ \text{eV}$. The only way to satisfy both conservation laws is to emit a photon that carries away both the excess energy and some momentum.

3. A single photon cannot do both jobs perfectly.

   Suppose the electron and proton capture each other and emit one photon. Energy conservation:

Momentum conservation:

The photon's momentum is $E_\gamma / c$. For a typical capture, the binding energy is $\sim 13.6\ \text{eV}$, so $E_\gamma \approx 13.6\ \text{eV}$. The photon's momentum is then $13.6\ \text{eV}/c$, which is very small. But the initial momenta of the electron and proton can be much larger (even for thermal energies, $k_B T \sim 0.025\ \text{eV}$ gives momenta of order $\sqrt{2mK} \sim \text{a few eV}/c$). The atom's recoil momentum must balance the difference, but the atom is heavy — its recoil kinetic energy is $p^2/(2M) \sim ( \text{a few eV}/c)^2 / (2 \times 10^9\ \text{eV}/c^2) \sim 10^{-9}\ \text{eV}$, negligible. So the photon cannot carry away enough momentum to balance the initial momenta unless the initial kinetic energies are extremely small.

In practice, for free particles at any reasonable temperature, a single photon cannot simultaneously conserve both energy and momentum. This is why free electron-proton capture requires a third body (like another photon or a nearby atom) to take up the slack.

4. Angular momentum is not the issue. The electron and proton can have any relative angular momentum. The hydrogen atom's ground state has zero orbital angular momentum ($l=0$), but the spins can combine to give either a singlet or triplet state. Angular momentum can be conserved by emitting a photon that carries away one unit of angular momentum (it is a spin-1 particle). So angular momentum conservation does not forbid capture — it just determines the photon's polarisation. Option (D) is false. …

Method: Pattern-Match to the "Two-Body Capture Needs a Third Body" Family of Problems

This question belongs to a recognisable class of physics problems — recognising the class lets you skip re-deriving the momentum-conservation argument from first principles and instead apply a known general principle.

Steps

Step 1: Recognise the general theorem

A rule worth remembering: a free particle cannot be captured into a bound state by emitting a single photon and nothing else, in general, because a single photon cannot simultaneously satisfy both energy conservation and momentum conservation for generic (nonzero) initial momenta. This is the same underlying reason a completely free electron cannot absorb a photon at all (the photoelectric effect always needs a bound electron, or Compton scattering needs the electron to remain free with a recoil angle) — it's a standard "two conservation laws, one massless quantum" mismatch that shows up throughout atomic and particle physics.

Step 2: Confirm which conservation law is actually doing the work

Write down both conservation statements symbolically:

$\text{Energy: } K_i = E_{\text{bound}} + K_f + E_\gamma, \qquad \text{Momentum: } \vec p_i = \vec p_f + \vec p_\gamma$ …