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Gamma radiation follows the inverse square law, I understand this as "double the distance, quarter the intensity"

So if you have a gamma source, at the source (distance = 0), the Intensity is $I_0$, and say at distance = 1, the Intensity is $\frac{I_0}{2}$ (You can't work this out just from the fact it follows the inverse square law right? You'd need the constant?)

So at distance = 2, while the intensity be a quarter of the original intensity so $\frac{I_0}{4}$ or a quarter of the intensity at the distance(1) that was doubled, so $\frac{I_0}{8}$?

I ask because I think this graph, which shows intensity of gamma radiation vs distance according the inverse square law, is wrong?

enter image description here

(also I don't see how it gets from 3x to $\frac{I_0}{8}$ because $3^2=9$)

Jonathan.
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1 Answers1

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This isn't the intensity as a function of distance from a point source in open space. It's intensity as a function of penetration through shielding. They're defining $x$ as the amount of material that a gamma ray has a probability of 1/2 of penetrating. Independent probabilities multiply, so the probability of penetrating three such thicknesses is $(1/2)^3$. The label on the horizontal axis probably means "thickness in units of meters."

  • Ahh, thanks. This graph is in the same page as inverse square law but the description is on the page before. But that brings up a question I would have thought that it would still follow the inverse square law (assuming the material is uniform) just with a different constant? – Jonathan. Jun 13 '13 at 22:56
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    @Jonathan. The inverse-square law follows from propagating in 3 dimensions away from a point, since the area of a sphere around the source grows as the square of the radius. This problem is assuming a planar source propagating in 1D, so the cross section of the beam has a constant area. This is the case whenever the source is large and/or far away compared to the penetration distance being observed. –  Jun 14 '13 at 00:10