determined using the cross section and number density of scatterers. From the scattering cross section of a particle, $\sigma$, and number density of the scatterers, $\rho$, the mean free path $l$ is determined as $$\begin{aligned}
l={1 \over \sigma \rho }.\end{aligned}$$ The mean free paths of various particles in matter are computed easily.

### Comparison of energy widths 

Because the particle state is defined using one wave function within the mean free path, $l$, this state has a finite uncertainty of the momentum, $\delta p$, $$\begin{aligned}
\delta p=\frac{\hbar}{l}.\end{aligned}$$ This finite uncertainty of the momentum leads to a finite uncertainty of the energy for the nonrelativistic particle of mass $m$, $$\begin{aligned}
\delta E={\delta p^2 \over 2m}=v\delta p=v\frac{\hbar}{l}.
\label{delta-energy-1}\end{aligned}$$ Thus, the particle that has a mean free path $l$ has the energy width $v{\hbar \over l}$.

The uncertainty of the energy of a wave packet is found from the momentum width Eq. $(\ref{parameter-sigma})$, $$\begin{aligned}
\delta E={\delta p^2 \over 2m}=v\delta p=v{\hbar \over \sqrt{ 2}\gamma}.
\label{delta-energy-2}\end{aligned}$$ Comparing two energy widths, Eqs. $(\ref{delta-energy-1} )$ and $(\ref{delta-energy-2} )$, we have the wave packet parameter $\gamma$, $$\begin{aligned}
\gamma= {l \over \sqrt 2}.\end{aligned}$$ Thus, the wave packet size $\gamma$ is determined using the mean free path $l$.

When this 