]> Nuclear Equation using MathML ${\mathrm{&CapitalIota;}}_{0}\approx \frac{{R}_{\Lambda }}{4\Pi {D}^{2}}{e}^{-ΜD}\phantom{\rule{3em}{0ex}}=\mathrm{&FreakedSmiley;}$

I0 is the radiation dose received at a distance D from a nuclear explosion. ${R}_{\Lambda }\phantom{\rule{0.3em}{0ex}}\text{is the initial total radiation release, and}\phantom{\rule{0.3em}{0ex}}Μ\phantom{\rule{0.3em}{0ex}}\text{is the attenuation factor in air. (e and}\phantom{\rule{0.3em}{0ex}}\Pi$ are mathematical constants). Ref: "The Effects Of Nuclear Weapons", US Dept. of Defense, April 1962, section 8.98.