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Determine the half value thickness, X(1/2) for lead where;

I = 0.4 MeV

u = 1.59cm^-1

2.

using;

I = Io e^-ux

3.

ln(I/Io) = ln(e) -ux

ln(2) / -1.59x10^-2 = X(1/2)

=> X(1/2) = -1.7523m

Is that right?

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# Homework Help: Half value thickness for lead

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