Q.Deuteron is a bound state of a neutron and a proton with a binding energy $B = 2.2\ \text{MeV}$. A $\gamma$-ray of energy $E$ is aimed at a deuteron nucleus to try to break it into a (neutron + proton) such that the n and p move in the direction of the incident $\gamma$-ray. If $E = B$, show that this cannot happen. Hence calculate how much bigger than $B$ must $E$ be for such a process to happen.

A photon of energy $E$ carries momentum $p_\gamma = E/c$, so breaking the deuteron must conserve both energy and momentum.

Why $E=B$ fails: if the photon supplied only the binding energy $B$, the neutron and proton would be left at rest (no kinetic energy), so their total momentum would be zero. But the photon brought momentum $E/c \neq 0$, which cannot disappear. Hence $E=B$ violates momentum conservation and the deuteron cannot split.

Threshold: the two nucleons must move forward together (total mass $\approx 2m$, with $m$ the nucleon mass) carrying momentum $E/c$, so their kinetic energy is

Energy conservation $E - B = K$ gives

$E - B = \frac{E^2}{4mc^2}.$ …

The physics

Binding energy alone is not the whole story. A photon of energy $E$ always carries momentum $p_\gamma = E/c$. When it is absorbed by the deuteron (initially at rest), the products must conserve both energy and momentum. Momentum conservation is what makes $E=B$ impossible.

Step 1 — Conservation laws

Let the photon move along $+x$ with energy $E$ and momentum $E/c$. The neutron and proton emerge in the $+x$ direction. Writing the nucleon mass as $m$ (so $m_n \approx m_p \approx m$, $mc^2 \approx 938\ \text{MeV}$) and the deuteron mass as $M_d = 2m - B/c^2$:

• Momentum: $\dfrac{E}{c} = p_n + p_p$
• Energy: $E + M_d c^2 = (\text{total energy of } n + p)$

Step 2 — Why $E=B$ is impossible

Suppose $E = B$. Absorbing the photon just supplies the energy needed to unbind the pair, so nothing is left over for kinetic energy — the neutron and proton would have to be produced at rest, giving zero total momentum. But the incoming photon carried momentum $E/c = B/c \neq 0$. Zero cannot equal $B/c$, so the two conservation laws cannot both hold. The reaction is forbidden at $E=B$.

Step 3 — Threshold energy

At threshold the nucleons move forward together with the same velocity (a composite of mass $\approx 2m$) carrying the whole momentum $E/c$. Their kinetic energy is …

Method: Threshold Energy for Photodisintegration — Applying Momentum Conservation Alongside Energy Conservation

This method applies to any "can a photon of energy $E$ break apart a bound system" question, where checking energy conservation alone is not enough — the photon's momentum must also be accounted for.

Steps

Step 1: Remember that a photon carries momentum as well as energy

A photon of energy $E$ has momentum $p_\gamma = E/c$. When it is absorbed by a nucleus initially at rest, both energy AND momentum must be conserved between the initial state (nucleus + photon) and the final state (the separated fragments).

Step 2: Test the naive case where the photon supplies exactly the binding energy

If $E$ equals the binding energy $B$ exactly, all of the photon's energy is used up unbinding the system, leaving no energy for the fragments' kinetic energy — so they would have to emerge at rest, with zero total momentum. But the incoming photon carried nonzero momentum $E/c$. Since momentum cannot simply vanish, this is a contradiction: $E=B$ is forbidden.

Step 3: Set up the true threshold condition

At threshold, momentum conservation requires the fragments to move off together (in the photon's original direction) carrying the full incoming momentum $E/c$. Treating them as one composite of total mass $M$, their kinetic energy is

Energy conservation then requires the photon energy to equal the binding energy plus this unavoidable kinetic energy: …