Introduction To Health Physics
Introduction To Health Physics
5th Edition
ISBN: 9780071835275
Author: Johnson, Thomas E. (thomas Edward), Cember, Herman.
Publisher: Mcgraw-hill Education,
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Chapter 9, Problem 9.23P
To determine

The radiation exposure rate.

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The linear attenuation coefficient for 2.0-MeV gamma rays in water is 4.9 m-1 and 52 m-1 in lead.  What thickness of water would give the same shielding for gamma rays as 15 mm of lead?
A small 10-gram source of cobalt-60 is in a vacuum. (a) What is the activity of the cobalt-60source in Bq? (b) What is the actual gamma-ray flux in cm2-s-1 at a point of measurement500 cm from the source due to the cobalt-60 emitted gamma-rays? (c) If the backgroundgamma-ray flux in the vicinity of the 10-g source is 7.8×107cm-2-s-1, what is the totalgamma-ray flux at the point of measurement, including the background? (d) What thicknessof a lead shield in cm would have to be placed between the source and the point ofmeasurement to reduce the total of the background plus the uncollided gamma-ray flux fromthe cobalt-60 source to 1.00×108cm-2-s-1 at that point? (I've attempted part a of the problem and don't know how to continue)
Tungsten has a linear attenuation coefficient of 30.5/cm for gamma rays with 150 keV of energy. If an unshielded Tc-99m syringe has a radiation exposure rate (I(0)) of 34 mR/hr, what will be the exposure rate if a 0.25-cm tungsten shield is placed around it? Using the calculation above, how much (what percent) is transmitted through the shield? What percent is attenuated?
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