Step-by-step Solution:

1. Understand the Physics of Beta Decay

The problem describes the beta decay of a neutron at rest into a proton, an electron, and an anti-neutrino: $n \to p + e^- + \bar{\nu}_e$ This is a three-body decay process. The total energy released in this decay is called the Q-value. This energy is conserved and is distributed among the kinetic energies of the decay products.

2. Apply the Law of Conservation of Energy

Since the initial neutron is at rest, its total energy is its rest mass energy, $m_n c^2$. After the decay, the total energy is the sum of the rest mass energies and kinetic energies of the products. $E_{initial} = m_n c^2$ $E_{final} = (m_p c^2 + K_p) + (m_e c^2 + K_e) + (m_{\bar{\nu}_e} c^2 + K_{\bar{\nu}_e})$ By energy conservation, $E_{initial} = E_{final}$. The problem states to assume the anti-neutrino is massless, so $m_{\bar{\nu}_e} = 0$. Its energy is purely kinetic, $E_{\bar{\nu}_e} = K_{\bar{\nu}_e}$. So, the energy conservation equation becomes: $m_n c^2 = m_p c^2 + K_p + m_e c^2 + K_e + K_{\bar{\nu}_e}$ The total kinetic energy shared by the products is the Q-value of the reaction: $Q = (m_n - m_p - m_e)c^2 = K_p + K_e + K_{\bar{\nu}_e}$ This Q-value is a constant for the decay process.

3. Relate Q-value to the Maximum Electron Kinetic Energy ($K_{e,max}$)

The problem states that the maximum kinetic energy of the electron is $K_{e,max} = 0.8 \times 10^6$ eV. The kinetic energy of the electron, $K_e$, will be maximum when the kinetic energies of the other two particles ($K_p$ and $K_{\bar{\nu}_e}$) are at their minimum. The minimum possible kinetic energy for the anti-neutrino is $K_{\bar{\nu}_e} \approx 0$. In this scenario, the decay is effectively a two-body problem involving the proton and electron. To conserve momentum (initially zero), the proton and electron must move in opposite directions with equal momentum magnitudes: $\vec{p}_p + \vec{p}_e = 0$. The proton, being much more massive than the electron ($m_p \approx 1836 \, m_e$), will have a very small recoil kinetic energy ($K_p = p_p^2 / (2m_p)$) compared to the electron. Therefore, $K_p$ is negligible. Thus, when $K_e$ is maximum, almost all the Q-value is carried by the electron: $Q \approx K_{e,max} = 0.8 \times 10^6 \text{ eV}$

4. Determine the Maximum Energy of the Anti-neutrino ($E_{\bar{\nu}_e, max}$)

The question asks for the maximum energy of the anti-neutrino. Since it is considered massless, its energy is equal to its kinetic energy, $E_{\bar{\nu}_e} = K_{\bar{\nu}_e}$. The energy of the anti-neutrino will be maximum when the kinetic energies of the other two particles ($K_p$ and $K_e$) are at their minimum. The minimum possible kinetic energy for the electron is $K_e \approx 0$. In this case, the energy conservation equation is $Q = K_p + K_{\bar{\nu}_e, max}$. Again, by momentum conservation, the proton and anti-neutrino must move in opposite directions with equal momentum magnitudes: $\vec{p}_p + \vec{p}_{\bar{\nu}_e} = 0$. The proton's recoil energy, $K_p$, will be very small compared to the energy of the massless anti-neutrino for the same momentum. Therefore, we can neglect $K_p$. Thus, when $K_{\bar{\nu}_e}$ is maximum, almost all the Q-value is carried by the anti-neutrino: $Q \approx K_{\bar{\nu}_e, max}$

5. Conclusion

From Step 3 and Step 4, we have: $Q \approx K_{e,max} \quad \text{and} \quad Q \approx K_{\bar{\nu}_e, max}$ This implies that the maximum kinetic energy of the electron is approximately equal to the maximum energy of the anti-neutrino. $K_{\bar{\nu}_e, max} \approx K_{e,max} = 0.8 \times 10^6 \text{ eV}$ Therefore, the maximum energy of the anti-neutrino is nearly $0.8 \times 10^6$ eV.

Comparing this result with the given options: A: Zero - Incorrect. B: Much less than $0.8 \times 10^6$ eV - Incorrect. C: Nearly $0.8 \times 10^6$ eV - Correct. D: much larger than $0.8 \times 10^6$ eV - Incorrect, as it violates energy conservation.