Maximum Energy of Anti Neutrino in Beta Decay Process

In the three-body decay of a stationary neutron, represented by: $n \rightarrow p + e^- + \bar{\nu}_e$

The conservation of total energy yields the available disintegration energy ($Q$-value) of the decay as: $Q = K_p + K_e + E_{\bar{\nu}_e}$

where:

• $K_p$ is the recoil kinetic energy of the proton,
• $K_e$ is the kinetic energy of the emitted electron,
• $E_{\bar{\nu}_e}$ is the energy carried away by the massless anti-neutrino.

Because the mass of the proton is much larger than the masses of the electron and anti-neutrino ($m_p \gg m_e$), the momentum conservation implies that the proton receives a negligible fraction of the total kinetic energy ($K_p \approx 0$). Consequently, the total energy $Q$ is almost completely distributed between the electron and the anti-neutrino: $K_e + E_{\bar{\nu}_e} \approx Q$

1. Maximum Electron Kinetic Energy: When the energy carried by the anti-neutrino is at its minimum ($E_{\bar{\nu}_e} \to 0$), the electron achieves its maximum kinetic energy: $(K_e)_{\text{max}} \approx Q = 0.8 \times 10^6\text{ eV}$

2. Maximum Anti-Neutrino Energy: Similarly, when the electron is emitted with negligible kinetic energy ($K_e \to 0$), the anti-neutrino carries away the maximum possible energy: $(E_{\bar{\nu}_e})_{\text{max}} \approx Q \approx 0.8 \times 10^6\text{ eV}$

Thus, the theoretical maximum energy carried away by the anti-neutrino is approximately $0.8 \times 10^6\text{ eV}$.

Correct Option: C