N1. Why Are There No Electrons In The Nucleus?

Before the discovery of the neutron in 1932, if nuclei were to be described in terms of known particles at all, it had to be in terms of protons and electrons. No other building blocks were known. An alpha particle, for example, could be described as a closely bound structure of four protons and two electrons. The four protons contributed the necessary mass; the two electrons canceled the excess charge. Similarly, a nucleus of uranium 238 could be assumed to contain 238 protons and 146 electrons. Quantum mechanics raised two serious objections against this view of nuclear composition. First, to hold an electron within the confines of a nuclear volume requires an energy larger than is actually associated with nuclear binding. Let me explain this with an illustrative calculation. If, for instance, an electron is held within the “container” of a helium nucleus, its wave amplitude must undergo at least one half cycle of oscillation in a distance of about 0.5 × 10^–14 m. This means that its wavelength can be at most 10^–14 m. Then the de Broglie equation reveals its least possible momentum:

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:

v_classical = 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.

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