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Radiation Saftey Problem

1. Homework Statement

In order to be granted tenure, you are required to spend the night on a mattress below which has been placed a highly radioactive pea. You may choose either a solid lead mattress which is 3cm thick or a cotton mattress (considered to be made of carbon of density 0.10 g/cm^3) which is 0.5 m thick. Assuming the pea has been doped with 65Zn, calculate the relative amount of protection afforded you by each mattress- i.e. the ratio between the doses you get in each case. Assume the thickness of the mattress equals the distance from the pea to the nearest spot on your skin (in the small of your back assuming you sleep on your back). Also assume most of th gammas which Compton scatter won't reach you, so you can use the sum of absorption and scattering cross sections. Calculate you are most concerned about reducing the dose to the most exposed cubic centimeter of your body.

2. Homework Equations

I know 65Zn=.330188 Roetgens/hr 1m from 1curie
density of Zn=6.92 g/cm^3

I know the amount of radiation you receive is proportional to the amount of time spent near the source and for distance is proportional to the inverse square law. I also know shielding protects against radiation, but I'm not sure what sort of proportionality I should use.

I know the depth of penetration from radiation is 1/absorption coefficient. However I do not know how to get the absorption coefficient.

I've also worked out Planck's Radiation Formula for the pea with radius = 1cm.
P=alphaAT^4=(5.67*10^8)(4pi)(294)^4=5323 W

So I guess what I need help on is where to look next.
 

Answers and Replies

454
0
You need to know the energy of the Gamma rays emitted. Absorption coefficients are tabulated here:http://physics.nist.gov/PhysRefData/XrayMassCoef/tab3.html" [Broken]

you want the attenuation coefficient, wich includes scattering reactions. See the links at the bottom of that page. These have to be multiplied by the density of the material to get the absorption depth.

Planck's radiation formula has nothing to do with nuclear reactions
 
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