Solving for Thickness of Lead to Reduce Count Rate to 50

In summary, the count-rate from a gamma source was measured to be 1000 counts per minute, but when 1.0 cm of lead was placed between the source and the detector, the count rate was reduced to 100 counts per minute. To further reduce the count-rate to 50 counts per minute, an additional thickness of 3 mm of lead would be needed. This can be calculated using the equation C= Cinital e^-μx, where C is the final count-rate, Cinital is the initial count-rate, μ is the attenuation coefficient, and x is the thickness of the material.
  • #1
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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  • #2
Your answer 13 mm should be correct
 
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  • #3
Your 13mm is correct, but it asks "what additional thickness...". It took 10 mm to reduce 1000 to 100.
 
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  • #4
phyzguy said:
Your 13mm is correct, but it asks "what additional thickness...". It took 10 mm to reduce 1000 to 100.
thank you!
 

1. How does the thickness of lead affect the count rate?

The thickness of lead directly affects the count rate by reducing the number of particles that can pass through it. As the thickness increases, more particles are absorbed, resulting in a decrease in the count rate.

2. What is the relationship between the lead thickness and the count rate?

The relationship between lead thickness and count rate is inversely proportional. This means that as the thickness of lead increases, the count rate decreases and vice versa.

3. How do you determine the appropriate thickness of lead to reduce the count rate to 50?

The appropriate thickness of lead can be determined by using a mathematical formula that takes into account the initial count rate and the desired reduced count rate. The formula is: t = -ln(0.5) / μ, where t is the thickness of lead and μ is the linear attenuation coefficient.

4. Can other materials be used instead of lead to reduce the count rate?

Yes, other materials with high linear attenuation coefficients can also be used to reduce the count rate. Examples include concrete, steel, and tungsten.

5. Is there a minimum thickness of lead required to reduce the count rate to 50?

The minimum thickness of lead required to reduce the count rate to 50 will depend on the initial count rate and the material's linear attenuation coefficient. However, in most cases, a thickness of 1-2 cm of lead is sufficient to reduce the count rate to 50.

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