As I know:

  • $E_\mathrm{p} = \pu{938.272 MeV}, \quad E_\mathrm{e} = \pu{0.511 MeV}, \quad E_\mathrm{n} = \pu{939.565 MeV}$
  • $E_\mathrm{n} − (E_\mathrm{p} + E_\mathrm{e}) = \pu{0.782 MeV}$
  • $M_\mathrm{p} = \pu{1.6727 x 10^-24 g}, \quad M_\mathrm{e} = \pu{9.110E-28 g}, \quad M_\mathrm{n} = \pu{1.6750E-24 g}$

Furthermore, following reaction also happens:

$\ce{p+ + e− -> n + ν}$

So i was wondering if this is a combination reaction at atomic scale and if the neutron is really a highly compressed form of hydrogen atom? As mass difference is negligible and higher energy of neutron can be regarded as the moving energies of electron(i.e. rotational, kinetic and others) are locked down into potential energy and charge of electron and proton neutralizing eachother?

  • $\begingroup$ While technically true, it does not make much sense to think of it this way. Look at everything around you (people, buildings, ground, vegetation, etc.) Now compress it with the same force. Everything will turn into neutrons. Are neutrons really a highly compressed form of all that? $\endgroup$ – Ivan Neretin Jun 19 '18 at 8:16
  • 1
    $\begingroup$ No, it seems not as the neutron can decay into a proton an electron and an anti-neutrino, and the proton can decay into a neutron a positron and an electron neutrino. Thus there are extra particles involved than just the proton and electron. $\endgroup$ – porphyrin Jun 19 '18 at 8:29
  • $\begingroup$ General thoughts along those lines were pretty common between the time of the identification of deuterium and the discovery of the neutron. But, nuclear physics makes you look at the quark constituents and that takes you into a completely different space, far from chemistry. So, the neutron is not a compressed form of hydrogen. $\endgroup$ – Jon Custer Jun 19 '18 at 13:43

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