18 | January | 2016 | Nuclear Physics 101

Mean free path of a nucleon inside a nucleus

January 18, 2016

The mean free path is the average distance between 2 collisions. We simply copy the things in Bohr and Mottelson, Bertulani and Banielewicz, and John Lilley in here.

Since a nucleus has finite size, the mean free path can be view as transparency of nucleon scattering. Since the total cross section is smaller for higher energy, the mean free path is proportional to energy.

The wave number $K$ under a complex optical potential $V+iW$ is

$K=\sqrt{\frac{2m}{\hbar^2}(E-V-iW)}=k_r+ik_i$ .

There are two solutions, one has imaginary $k_r$ ,

$k_r=\frac{\sqrt{m}}{\hbar}\sqrt{(E-V)+\sqrt{(E-V)^2+W^2}}$

$k_i=\frac{m}{\hbar^2}\frac{W}{k_r}=\frac{1}{2\lambda}$

where $m$ is the nucleon mass, $E$ is the incident energy, $v$ is the velocity inside the nucleus, and $\lambda$ is the mean free path,

$\lambda=\frac{\hbar}{2W\sqrt{m}}\sqrt{(E-V)+\sqrt{(E-V)^2+W^2}}\approx\frac{\hbar}{W\sqrt{2m}}\sqrt{(E-V)}$

Since the imaginary potential $W\approx-15-0.07E$ , $V\approx-39+0.11E$ from 150 MeV to 400 MeV. The calculated mean free path for proton is shown in here. (Take $\hbar = 197.33 [MeV\cdot fm/c]$ , proton mass $m=938.272 [MeV/c]$ , the $\lambda$ is in fm.)