Radioactivity and Shielding
I am working on an experiment of radioactivity for my
physics class and I found the absorption of radiation by matter using
aluminum foil and placing various numbers of sheets of it between the
gamma source and the Geiger-Muller tube, I found the counts per minute
using for different numbers of foils placed in between the source and the
tube. I graphed thickness vs. ln I (ln of cpm). I wanted to know What
the coefficient of linear absorption is for aluminum(the actual value so
I can compare it to my experimental)???? What would have been different
if another material was used instead of aluminum for absorption? I have
lnI = -ux + lnIo....would the coefficient of linear absorption been
different with different material?
When radiation is absorbed, one "piece" of the radiation is picked uo by an
atom or molecule of the absorbing material. It may be turned into heat, or
it may be re-emitted in another direction.
For gamma radiation, the "pieces" are photons. A large number are released
by the gamma source.
Some get stopped, while some pass through without hitting any aluminum
Different materials have different abilities to stop a gamma photon. Some
materials, such as lead and iron, are very good at absorbing a gamma photon.
Some materials, such as paper, are not very good at all. A sheet of lead
will absorb many more photons than will a sheet of paper with the same
thickness. Just as important as individual molecules is how tightly packed
the molecules are. Molecules that are very close together are more likely
to have a photon come into contact. The photon still may not be absorbed,
but it will have more molecules to get past.
For a VERY thin sheet of material, the coefficient of linear absorption is
(the portion of the intensity absorbed) divided by (the thickness). For
thicker sheets, you have to use the logorithmic relation.
I expect sheets of paper or plastic will give significantly smaller
coefficients than aluminum.
Dr. Ken Mellendorf
Illinois Central College
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Update: June 2012