# Velocities of Nucleons

by jaketodd
Tags: nucleons, velocities
 P: 3,014 Velocities of Nucleons Assume the nuclear matter (protons and neutrons as indistinguishable) forms a degenerate fermi gas. The fermi momentum is determined by: $$A = 2 \, \frac{4 \pi R^{3}}{3} \, \frac{4 \pi k^{3}_{F}} {3} \, \frac{1}{(2\pi)^{3}} = \frac{4}{9 \pi} (k_{F} R)^{3} \Rightarrow R = \left(\frac{9\pi}{4}\right)^{\frac{1}{3}} \frac{A^{1/3}}{k_{F}} = R_{0} \, A^{1/3}$$ $$R_{0} = \left(\frac{9\pi}{4}\right)^{\frac{1}{3}} \frac{1}{k_{F}} \Rightarrow k_{F} = \left(\frac{9\pi}{4}\right)^{\frac{1}{3}} R^{-1}_{0}$$ Using the emprical result $R_{0} = 1.2 \, \mathrm{fm}$, we get: $$k_{F} = 1.6 \, \mathrm{fm}^{-1}$$ The momentum corresponding to this is: $$p_{F} \, c = \hbar \, c \, k_{F} = 316 \, \mathrm{MeV}$$ which is one third of the rest energy of a proton (neutron). That is why one should use relativistic equation: $$p = m \, c \, \beta, \gamma, \ \gamma = (1 - \beta)^{-1/2}, \; \beta = v/c$$ Then, use the fact that you have a FD distribution in momentum space to find the distribution in velocity space. From this distribution you can find the most probable speed, the average speed and the root mean square speed, for example. A rough estimate, however is to simply use the fermi momentum: $$\beta_{F} = v_{F}/ c = \frac{p_{F}/(m \, c)}{\sqrt{1 + (p_{F}/(m \, c))^{2}}} = 0.32$$