How is the speed of nucleons in the nucleus measured?

If you shoot an electron or a proton at a nucleus at moderate energies (a few hundred $\mathrm{MeV}$ to a few $\mathrm{GeV}$) it will usually either bounce off the whole nucleus or break up the nucleus. But every once in a while (and this gets rarer and rarer the harder you throw it in) it will actually bounce off of a single nucleon. At the right energies this can happen without exciting the target nucleon (or the beam nucleon if you are using a proton beam), and still knock that nucleon right out of the parent nucleus without doing too much mischief on the way out.

These events are termed "quasi-elastic scattering" (not to be confused with the use of the same term in neutrino scattering).

The "quasi" is because there is some interaction between the beam particle and nucleus on the way in and out and some interaction between the scattered particle and the remnant-nucleus on the way out. But this is fairly modest and and can be computed in simulation.

It is the "elastic" part which we concentrate on. Elastic collisions between two objects are fully constrained by energy and momentum conservation. We know the initial energy and momentum of the beam particle, so if we measure the energy and momentum of the scattered particles we can deduce the energy and momentum of the target particle before it was hit.

Then all that remains is to computationaly remove the effects of the final-state interactions. Well, and we have to allow for the fact that a nuclear proton is slightly different from a free proton, and for that we rely on phenomenological models.

This is exactly the kind of data we took in my disseration experiment (our purpose was a little more subtle than just making the momentum measurement, but we got that for free). We took events like $A(e,e'p)$ using a $5$-$6 \,\mathrm{GeV}$ electron beam on protium, deutrium, carbon, and iron targets.

This figure appeared in my dissertation. It shows the processed result for squared momentum transfer of $3.3 \,\mathrm{GeV}^2$ on a carbon target.

• In the right-hand panel we plot the magnitude of the initial momentum of the struct particle in units of $\mathrm{GeV}/c$. You can see that the protons are mildly relativistic (recall that their mass is nearly $1 \,\mathrm{GeV}/c^2$).

• In the left-hand panel we plot the binding energy of the struck proton.

• We see the shell structure of the nucleus. The s-shell is responsible for the smaller, more-tightly bound energy hump and the non-zero density at the center of the momentum graph. The p-shell is responsible for the less tightly bound energy hump and the two-lobed structure of the momentum graph.

There are a variety of ways of getting this kind of information. The simplest way to get a rough estimate is just to use the size of the nucleus and the de Broglie relation. A moderate-sized nucleus has a diameter of roughly 7 fm, so the lowest-energy standing wave would have a wavelength of 14 fm along any given axis $x$. Plugging in to the de Broglie relation gives $v_x/c\approx 0.1$ in this example.

An even simpler variant of Ben's answer. Use the uncertainty principle. $\Delta p\ \Delta x \geq \hbar/2$.

If a nucleon is trapped in a box of order $\Delta x \sim 1.2\times 10^{-15} A^{1/3}$ m, where $A$ is the atomic mass, then its momentum must be of order $10^{15} A^{-1/3}\hbar$.

Inserting the mass of a nucleon gives a speed of order $0.2 A^{-1/3}c$.

[Side note: All nuclei have roughly the same density.]