@heather This factor is usually ignored first, and the resulting stopping powers are given in MeV cm^2/g. You can multiply the results with the actual density later.
Also, note that anything with an actual number density in it is bound to have some ridiculously big numbers in it (of the order of Avogadro's number), which makes it a lot less helpful
having something in terms of the mass density is a lot more helpful
@0celo7some guy told my friend that he didn't believe that the ideas from QM were correct. The guy was a computer engineer. So my friend told him that without the theory of QM we would not have been able to create a classical computer bit.
well, it'll end up being m^2*MeV which I suppose might make sense - I guess it represents the distance the electron travels at a certain energy before energy lost due to bremsstrahlung radiation becomes comparable to energy lost due to ionization, maybe?
@0celo7 Compare to N_2 and O_2, which have rotational excited states at room temperature, but different rotational energy spectra because the rotor masses are different?
@rob My homework wants to know if the $\Delta U$ in some process changes if I replace neon with argon. I think it's a "yes" because they have different heat capacities
@0celo7 Right. Because in that ratio the mass of the atom cancels out, and the ideal gas equation $pV = NkT$ treats pressure and volume the same way as far as storing energy in the gas.
@heather OK, so we have $-\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=\frac{(e^2)^2}{10(m_0c^2)^2}NE\:Z^2=(8.3\times 10^{-31}\:\mathrm m^2)NE\:Z^2$
@heather This factor is usually ignored first, and the resulting stopping powers are given in MeV cm^2/g. You can multiply the results with the actual density later.
let's put that into $-\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=\frac{(e^2)^2}{10(m_0c^2)^2}NE\:Z^2=(8.3\times 10^{-31}\:\mathrm m^2)NE\:Z^2$
so you have $-\frac1\rho\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=\frac{(e^2)^2}{10(m_0c^2)^2}\frac N\rho E\:Z^2=(2.44\times 10^{-6}\:\mathrm{m^2/kg}) E\:Z^2$, then
we had $-\frac1\rho\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=\frac{(e^2)^2}{10(m_0c^2)^2}\frac N\rho E\:Z^2=(2.44\times 10^{-6}\:\mathrm{m^2/kg}) E\:Z^2$
and we had $-\frac1\rho\left(\frac{\mathrm dE}{\mathrm dx}\right)_\mathrm r=\frac{(e^2)^2}{10(m_0c^2)^2}\frac N\rho E\:Z^2=(2.44\times 10^{-6}\:\mathrm{m^2/kg}) E\:Z^2=0.14\:\mathrm{MeV\:m^2/kg}$