1. We check each option using conservation of charge, energy, and momentum.
   A process is allowed only if all conserved quantities can be satisfied simultaneously.

Given masses: $m_p=1.0072\,u,\quad m_n=1.0087\,u,\quad m_e=0.000548\,u,\quad m_{\bar\nu}=0,\quad m_d=2.0141\,u$

Also, the mass of a deuterium atom is $m(\text{deuterium atom})=m_d+m_e=2.0141+0.000548=2.014648\,u$

2. Option A: $n+n \to$ deuterium atom

Reaction: $n+n \to d+e^-$ (since deuterium atom means deuteron + bound electron)

(i) Charge conservation

Initial charge: $0+0=0$ Final charge: $+1+(-1)=0$ So charge is okay.

(ii) Energy conservation

Initial mass: $2m_n=2(1.0087)=2.0174\,u$ Final mass: $m_d+m_e=2.0141+0.000548=2.014648\,u$ Mass difference: $\Delta m=2.0174-2.014648=0.002752\,u>0$ So energy could in principle be released.

(iii) Momentum conservation

This is a two-body initial to one composite final object process if treated as just forming a deuterium atom. In the center-of-mass frame, initial total momentum can be zero, but then the final single particle must also have zero momentum. That would force its total energy to be just its rest energy, while initial energy is larger. The excess energy has nowhere to go.

So with only one final object, simultaneous energy and momentum conservation is impossible unless initial total energy exactly equals final rest mass, which it does not.

Hence A is not allowed.

3. Option B: $n+p \to d+\gamma$

This is radiative capture.

(i) Charge conservation

Initial charge: $0+(+1)=+1$ Final charge: $d\text{ has charge }+1,\quad \gamma\text{ has }0$ So charge is conserved.

(ii) Energy check

Initial mass: $m_n+m_p=1.0087+1.0072=2.0159\,u$ Final rest mass (excluding photon energy): $m_d=2.0141\,u$ Mass difference: $\Delta m=2.0159-2.0141=0.0018\,u>0$ So the released energy can be carried away by the photon.

(iii) Momentum conservation

Because there are two final particles ($d$ and $\gamma$), momentum can be conserved: in the center-of-mass frame, the deuteron and photon recoil with equal and opposite momenta.

Therefore B is allowed.

4. Option C: $p \to n+e^++\bar\nu$

(i) Charge conservation

Initial charge: $+1$ Final charge: $0+(+1)+0=+1$ Charge is conserved.

(ii) Energy conservation

Initial mass: $m_p=1.0072\,u$ Final minimum mass: $m_n+m_e+m_{\bar\nu}=1.0087+0.000548+0=1.009248\,u$ Since $1.009248\,u > 1.0072\,u$ we need extra mass-energy, but none is available for a free proton decay.

Thus energy conservation fails.

Hence C is not allowed.

5. Option D: $e^++e^- \to \gamma$

(i) Charge conservation

Initial charge: $+1+(-1)=0$ Final charge: $0$ Charge is fine.

(ii) Momentum and energy conservation

A single photon has energy-momentum relation $E_\gamma = p_\gamma c$ In the center-of-mass frame of $e^+$ and $e^-$, initial total momentum can be zero. If only one photon is produced, final momentum would be nonzero for any real photon, which is impossible. A photon cannot have zero momentum unless it has zero energy, so one-photon annihilation in free space is forbidden.

Therefore D is not allowed.

(That is why free $e^+e^-$ annihilation normally produces two photons.)

6. Conclusion Only option B satisfies both energy and momentum conservation.

$\boxed{\text{B}}$