Answer to Question #118614 in Atomic and Nuclear Physics for dom

Question

A 2.0 mm thick sheet of shielding allows 80% of incident gamma radiation to pass through it.
How many layers of this shielding are required such that less than 30% of incident gamma rays to pass through all layers? (to 2 s.f)

Answer 1

We know that one layer allows 80% of incident radiation to pass and absorbs 100%-80% = 20% of radiation.

Let $X$ be the initial amount of radiation. After passing through one layer, 80% of $X$ (or $0.8X$ ) will remain. This amount of radiation falls onto the second layer, and after passing through it $0.8\cdot0.8 X = 0.64X$ will remain. Therefore, the amount of radiation remaining after passing through $N$ layers is $0.8^NX.$ Let us determine $N$ for which $0.8^NX \le 0.3X$ or $0.8^N\le0.3$ .

$0.8^5 = 0.32768 > 0.3, \;\; 0.8^6 = 0.262144 < 0.3.$ Therefore, the minimal number of layers is 6 and the total thickness will be $2.0\cdot6 = 1.2\,\mathrm{mm}.$