1. Key idea

In the decay $n \to p + e^- + \bar\nu_e,$ the electron gets a continuous range of kinetic energies because the available decay energy is shared among three particles.

The question asks: if the anti-neutrino has rest mass $m_\nu c^2 = 3\ \text{eV},$ what is the possible range of electron kinetic energy $K$?

2. Given information

For a massless anti-neutrino, the maximum electron kinetic energy is given as $K_{\max} = 0.8\times 10^6\ \text{eV}.$ This corresponds to the situation where the anti-neutrino carries negligible energy, and the proton only takes tiny recoil energy.

So the total energy available for electron kinetic energy plus neutrino energy is about $Q \approx 0.8\times 10^6\ \text{eV}.$

3. Effect of neutrino mass on maximum electron kinetic energy

If the anti-neutrino has mass, then at minimum it must carry its rest energy: $E_{\nu,\min} = m_\nu c^2 = 3\ \text{eV}.$

Hence the electron can no longer take the full $Q$ value. The new maximum electron kinetic energy becomes approximately $K_{\max}' = Q - m_\nu c^2 = 0.8\times 10^6 - 3\ \text{eV}.$

Since $3\ \text{eV} \ll 0.8\times 10^6\ \text{eV},$ this is practically still $K_{\max}' \approx 0.8\times 10^6\ \text{eV}.$

So the upper limit remains essentially unchanged at the level of the options.

4. Minimum electron kinetic energy

Can the electron have zero kinetic energy? Yes.

The available decay energy can be shared so that the electron is produced with arbitrarily small kinetic energy, while the proton and anti-neutrino carry the rest (subject to conservation laws).

The neutrino rest mass does not force the electron to have at least $3\ \text{eV}$ kinetic energy. The $3\ \text{eV}$ is simply the neutrino's rest energy requirement, not a lower bound on electron kinetic energy.

Therefore, $K_{\min} = 0.$

5. Range of electron kinetic energy

Thus the electron kinetic energy range is $0 \le K \lesssim 0.8\times 10^6\ \text{eV},$ which matches the option written as $0 \le K \le 0.8\times 10^6\ \text{eV}.$

So the correct option is A.

6. Check options

• A: $0 \le K \le 0.8\times 10^6\ \text{eV}$

  • Correct in the sense of the given options; strictly the upper limit is smaller by $3\ \text{eV}$, but that is negligible compared to $0.8\times 10^6\ \text{eV}$.

• B: $3\ \text{eV} \le K \le 0.8\times 10^6\ \text{eV}$

  • Incorrect. Electron kinetic energy need not be at least $3\ \text{eV}$.

• C: $3\ \text{eV} \le K \le 6\ \text{eV}$

  • Clearly incorrect.

• D: $0 \le K \le 6\ \text{eV}$

  • Incorrect. The upper limit remains of order $0.8\times 10^6\ \text{eV}$, not $6\ \text{eV}$.

7. Conclusion

The physically correct range is $\boxed{0 \le K \le 0.8\times 10^6\ \text{eV} \text{ (approximately)}}$ so the correct option should be A, not D.