p = h/λ = 6.63 x 10–34 kg m2/s / 10–14 m = 6.63 x 10–20 kg m/s
At this point I must proceed with caution. Is this momentum in the domain of relativistic mechanics or of classical mechanics? If I guess it to be classical, I can calculate the speed of the electron to be:
vclassical = p/m = 6.63 x 10–20 kg m/s / 9.11 x 10–31 kg = 7.3 x 1010 m/s
I have guessed wrong. This speed is 240 times greater than the speed of light, a physical impossibility. Therefore I must turn to a relativistic equation. A helpful one will be the equation connecting energy and momentum:
E2 = p2c2 + m2c4 or E2 = (pc)2 + (mc2)2 .
The quantity mc2, the rest energy of the electron, is, in electron volt units, mc2 = 0.511 MeV.
The quantity pc can be calculated for this example:
pc = (6.63 ×10–20 kg m/sec) ×(3 ×108 m/sec) = 1.99 ×10–11 J = 124 MeV .
(I have used the conversion factor, 1.6 × 10–19 J/eV.) Since pc is so much greater than mc2 (which is a mere 0.511 MeV), the second term in the energy-momentum equation above contributes very little, and the energy-momentum relation becomes, approximately, E = pc. In round numbers, then, an electron held within a helium nucleus would have a kinetic energy in excess of 100 MeV. So energetic an electron should in fact fly out of the nucleus, since the total binding energy holding all the particles together in the helium nucleus is only 28 MeV, or 7 MeV per nucleon. This was one difficulty with the idea of electrons in the nucleus.
Another difficulty concerns nuclear spin. According to the proton-electron model of nuclear composition, a nucleus of 7N14 contains 21 particles: 14 protons and 7 electrons. According to the proton-neutron model, the same nucleus contains only 14 particles: 7 protons and 7 neutrons. Since all the particles in question have one-half unit of spin, it makes an important difference whether the nucleus contains an odd or an even number of particles. If the number is odd, the total nuclear spin must be equal to an integer plus one half (in units of ħ). If the number is even, the nuclear spin must be integral. Early evidence that the nitrogen nucleus has integral spin was evidence against electrons in the nucleus. Since then, the spins of hundreds of nuclei have been determined. All are consistent with Heisenberg’s theory of neutron-proton composition.