 Solution

Define the positive mass collision stopping power as $S_{\mathrm{col}}/\rho =-(1/\rho )dT/dx$. For a projectile of charge $ze$, mass $M\gg m_e$, speed $v=\beta c$ and Lorentz factor $\Gamma =(1-{\beta }^2{)}^{-1/2}$, the Bethe stopping formula can be writtenHere $Z$ is the absorber atomic number, $A$ its molar mass, $I$ its mean excitation energy, $r_e=e^2/(4\pi {ϵ}_0{m}_e{c}^2)$, and $N_A$ the Avogadro constant. AlsoThe optional $\delta$ and $C/Z$ represent density and shell corrections. For a very heavy projectile $T_{\max }\simeq 2m_e{c}^2{\beta }^2{\Gamma }^2$, and the bracket reduces to $\ln (2m_e{c}^2{\beta }^2{\Gamma }^2/I)-{\beta }^2$ when those corrections are neglected. If $T$ denotes remaining projectile kinetic energy, the printed $dT/(\rho dx)$ is negative; the conventional positive stopping power has the minus sign above.^[ref]