I'm reading over one of Andre Gsponer's papers on nuclear weapons. I'm confused by section 3.2, particularly formulas 3 and 4. (3) states:

$$ \eta_\mathrm{fission} = \frac{\rho R - \omega_c}{\rho R} $$

I believe $\rho R$ term roughly equates to "number of atoms between the center and outside", but the $\omega_c$ term is an areal density. I am confused about what this means.

The following text states:

$\omega_c$ is called the “critical fast-neutron-opacity” ($\omega_c \approx 100$ g/cm² for ${}^{239}\mathrm{Pu}$, and $160$ g/cm² for ${}^{235}\mathrm U$)".

I assume from this terminology that it is a measurement of the point where the fuel becomes opaque to neutrons and thus you have an ~100% chance of any particular neutron causing another reaction. Uncompressed ${}^{235}\mathrm U$ is about $19$ g/cm³, so this seems to be suggesting that a chain reaction is only possible in compressed fuel, which is obviously not true. I think I am missing the meaning of $\rho R$.

(4) talks about fusion. I believe the term B, "$B \approx 6$ g/cm² for DT , and $17$ g/cm² for ${}^6Li_2DT$", is referring to the density at which the alphas from the DT reaction will be fully captured within the fuel mass and thus self-heat. But again, I am not entirely sure I understand the meaning of $\rho R$ here either.

  • $\begingroup$ Areal densities are used (either grams/cm2 or atoms/cm2) when considering scattering interactions from a flux of particles (measured in particles/cm2). This normalizes out any mention of actual material densities - the material could be a gas or a solid, you just want to know how many atoms an incident particle passes by before it might scatter. An actual mean free path, on the other hand, requires knowing the density. $\endgroup$
    – Jon Custer
    Apr 15 at 14:25


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.