Solving for Thickness of Lead to Reduce Count Rate to 50

AI Thread Summary
The discussion revolves around calculating the additional thickness of lead needed to reduce the gamma count rate from 100 counts per minute to 50 counts per minute. The initial count rate is 1000 counts per minute, and placing 1.0 cm of lead reduces it to 100 counts per minute. The attenuation coefficient was calculated incorrectly, leading to confusion about the additional thickness required. The correct calculation indicates that while 13 mm is needed to reach 50 counts per minute, only an additional 3 mm of lead is required beyond the initial 10 mm that reduced the count to 100. Clarification emphasizes understanding the question's focus on "additional thickness" rather than total thickness.
gungo
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Homework Statement


The count-rate from a gamma source is measured to be 1000 counts per minute. When 1.0 cm of lead is placed between the source and the detector, the count rate is reduced to 100 counts per minute. What additional thickness of lead would have reduced the count-rate to 50 counts per minute?

Homework Equations


C= Cinital e^-μx

The Attempt at a Solution


I'm not really sure how to approach the question I tried using the first equation to find the attenuation coeffecient by subbing in 100= 1000 e^-μ(0.01) and getting 230.26. And then I used that to find x by doing 50= 1000e^-230.26x and I got 13 mm, but the answer is 3 mm. Not sure where I'm going wrong or if I am even using the right formula...
 
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Your answer 13 mm should be correct
 
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Your 13mm is correct, but it asks "what additional thickness...". It took 10 mm to reduce 1000 to 100.
 
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phyzguy said:
Your 13mm is correct, but it asks "what additional thickness...". It took 10 mm to reduce 1000 to 100.
thank you!
 
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