Q.In pair annihilation, an electron and a positron destroy each other to produce gamma radiation. How is the momentum conserved?

The key idea is that the two gamma-ray photons are emitted in opposite directions with equal energy, so their vector momenta cancel exactly.

1. Before annihilation, the electron and positron are nearly at rest (thermal energies are negligible). Their total momentum is effectively zero.
2. After annihilation, two photons are produced. Each photon carries momentum $p = E/c$, where $E$ is its energy. …

The question touches on a beautiful moment in physics — when matter turns entirely into energy, yet the most fundamental laws of mechanics still hold. Let's build the understanding from the ground up.

The core idea: momentum doesn't vanish

When an electron and a positron meet, they annihilate. Their rest mass (about $0.511 \,\text{MeV}/c^2$ each) converts into pure energy in the form of gamma-ray photons. But momentum is a conserved quantity in every interaction — it cannot simply disappear. So the real question is: how does the momentum of the initial particles get carried away by the outgoing radiation?

The answer lies in the fact that two photons are produced, not one.

Step-by-step reasoning

1. Start with the initial conditions.

Before annihilation, the electron and positron are typically moving slowly — often they are nearly at rest relative to each other (e.g., in thermal motion or bound in a material). Their total momentum is therefore very close to zero. Even if they have some small kinetic energy, the net momentum vector is essentially zero.

2. Why can't a single photon be emitted?

If only one photon were produced, it would have to carry away all the energy ($E = 2 \times 0.511 \,\text{MeV} = 1.022 \,\text{MeV}$). But a photon has momentum $p = E/c$. So a single photon would have momentum $p = 1.022 \,\text{MeV}/c$ in some direction. That would violate conservation of momentum, because the initial momentum was zero. A single-photon annihilation is impossible in free space — it would require a third body (like a nucleus) to absorb the excess momentum.

3. The two-photon solution.

The simplest way to conserve both energy and momentum is to produce two photons of equal energy ($0.511 \,\text{MeV}$ each) travelling in exactly opposite directions. Their momenta are equal in magnitude but opposite in direction:

• Photon 1: momentum $+\vec{p}$
• Photon 2: momentum $-\vec{p}$

The vector sum is $\vec{p} + (-\vec{p}) = 0$, matching the initial zero momentum.

4. Energy is also conserved.

Each photon carries $0.511 \,\text{MeV}$ of energy, so total energy is $1.022 \,\text{MeV}$, exactly equal to the rest energy of the electron-positron pair ($2 m_e c^2$). No kinetic energy is left over — the annihilation is complete. …

Method: Applying Momentum Conservation to Particle–Antiparticle Annihilation

Use this whenever a question asks how momentum (or energy) is conserved when a particle-antiparticle pair converts entirely into radiation.

Steps

Step 1: Establish the initial total momentum

If the particle and antiparticle are (approximately) at rest before annihilating, their combined initial momentum is essentially zero:

Momentum conservation then requires the total momentum of whatever is produced to also equal zero.

Step 2: Show why a single photon cannot satisfy this

A photon of energy $E_\gamma$ necessarily carries momentum $p_\gamma = E_\gamma/c \ne 0$ in some direction. If only one photon were emitted, it alone would carry away nonzero momentum — violating the requirement from Step 1. So single-photon annihilation in free space is forbidden.

Step 3: Introduce a second photon to balance momentum

The minimum fix is to produce two photons, emitted in exactly opposite directions with equal magnitude of momentum:

This restores the total momentum to zero, matching the initial state. …