Solving for Thickness of Lead to Reduce Count Rate to 50

Click For Summary
SUMMARY

The discussion centers on calculating the additional thickness of lead required 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 1.0 cm of lead reduces this to 100 counts per minute. The correct additional thickness of lead needed is 3 mm, following the attenuation formula C = Cinital e^-μx. The confusion arose from misinterpreting the question regarding the additional thickness required after the initial reduction.

PREREQUISITES
  • Understanding of gamma radiation and its interaction with matter
  • Familiarity with the attenuation formula C = Cinital e^-μx
  • Knowledge of the concept of linear attenuation coefficient (μ)
  • Basic algebra skills for solving exponential equations
NEXT STEPS
  • Study the concept of linear attenuation coefficients in various materials
  • Learn how to apply the exponential attenuation formula in different scenarios
  • Research the properties of gamma radiation and its shielding requirements
  • Explore practical applications of radiation shielding in medical and industrial contexts
USEFUL FOR

Students in physics or engineering, radiation safety professionals, and anyone involved in radiation shielding calculations will benefit from this discussion.

gungo
Messages
27
Reaction score
1

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...
 
Physics news on Phys.org
Your answer 13 mm should be correct
 
Last edited:
  • Like
Likes   Reactions: gungo
Your 13mm is correct, but it asks "what additional thickness...". It took 10 mm to reduce 1000 to 100.
 
  • Like
Likes   Reactions: gungo
phyzguy said:
Your 13mm is correct, but it asks "what additional thickness...". It took 10 mm to reduce 1000 to 100.
thank you!
 

Similar threads

Replies
4
Views
2K
Replies
3
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K