A free neutron decays into a proton but a free proton does not decay into neutron. This is because

A free neutron is unstable and undergoes $$\beta^{-}$$-decay:$$n \rightarrow p^{+} + e^{-} + \bar\nu_e$$

For any decay to occur spontaneously, the total rest-mass (plus kinetic energy) of the products must be ≤ the rest-mass of the parent particle, otherwise the extra energy required cannot be supplied.

Rest-mass (in $$\mathrm{MeV}/c^{2}$$):
neutron $$m_n = 939.565$$
proton $$m_p = 938.272$$
electron $$m_e = 0.511$$

Total rest-mass of the decay products:
$$m_p + m_e = 938.272 + 0.511 = 938.783$$

Mass difference:
$$\Delta m = m_n - (m_p + m_e) = 939.565 - 938.783 = 0.782\;\mathrm{MeV}/c^{2}$$

The positive $$\Delta m$$ (≈ $$0.782\;\mathrm{MeV}$$) appears as kinetic energy of the proton, electron and antineutrino, so the decay is energetically allowed.

Now consider the reverse process (a free proton decaying into a neutron). The proposed products must at least include a neutron and a positron (to conserve charge) and a neutrino:$$p^{+} \rightarrow n + e^{+} + \nu_e$$

Rest-mass required on the right:$$m_n + m_{e^{+}} = 939.565 + 0.511 = 940.076\;\mathrm{MeV}/c^{2}$$

The initial rest-mass on the left is only $$m_p = 938.272\;\mathrm{MeV}/c^{2}$$, which is $$1.804\;\mathrm{MeV}/c^{2}$$ smaller. Energy conservation would therefore be violated; there is no external source to supply this extra energy, so the decay of a free proton is forbidden.

Hence, the key reason is that the neutron’s rest-mass is larger than the proton’s. Option D matches this explanation.

Final Answer: Option D