Absorption Of γ - Radiation in Lead/Absorption coefficient

AI Thread Summary
The discussion focuses on the absorption of gamma radiation in lead and the calculation of the absorption coefficient (μ) using experimental data. The user applied the equations I = I0 e-μd and C(d) = C0 e-μd to derive values for μ, resulting in estimates of approximately -0.0116 and -0.0109 from different graph analyses. There is confusion regarding the correct application of the equations and the interpretation of the background count in calculations. The user seeks clarification on how to accurately compute μ and reconcile the values obtained from the equations with the experimental data. Overall, the thread highlights the challenges faced in understanding and applying the mathematical concepts related to radiation absorption.
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EXPERIMENT ABSORPTION OF γ - RADIATION IN LEAD AND CALCULATION OF ABSORPTION COEFFICIENT µ

Using equation (3) / slope of graph lnC vs d - estimate µ. This should be in units of per meter (m-1 ).

The formula in the analysis are as follows :
1. I = I0 e-μd

2. C(d) = C0 e-μd

3. lnC = lnC0 -μd
You may ignore the decimal point at the RANGE=10 setting on the scaler/timer ! so the count then turns out to be whole numbers .

Data (where d is the thickness of the lead sheets):
  • d(m) = 0.0121 ,0.0063 ,0.0025 ,0.0184 ,0.0146 ,0.008
  • Count/min = 636 ,1050 ,1474 ,424 ,454 ,700
  • C (counts/min) - background count = 28
  • Find ln C
For the graph i subbed in the values of C to lnC and plotted it vs d(m) ,the result i got as for the line eq. was
y = -0.0116x +0.0886 in excell with the tread line . I hope the axis were right :P .
So μ≈ -0.0116

Using the equation was hard for me as I'm not sure which component is which , i feel like what I am doing is awfully wrong. I tried to work out μ by subbing in values from the table to C(d) and C0 equation 2 i did the same for the 3 equation and tried to solve with rules of logs , taking C0 as background count of 28 .

I would appreciate any explanation so i can do the calculation and see what happens .
 
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So =

d (m) = 0.0025, 0.0063 , 0.0089 , 0.0121 , 0.0146 ,0.0184
count/min= 1.474, 1.050 , 0.700 , 0.636, 0.454, 0.424
C (counts/min) - backgnd count = 1.446 , 1.022 , 0.672 ,0.608 , 0.426 , 0.396 ( that is subtracting 0.028 from the counts )
ln C = 0.3688 , 0.02176 ,-0.3975 ,-0.4976 ,-0.8533 ,-0.9263 ( that is subbing in C backgnd for C )

graph yields : y = -0.0109x + 0.0065 hence μ = -0.0109

i tired to solve the c(d) in equation 2 with the info we have + the mu from graph .
I thought it should give me back the value from before we subtracted the backend (C counts/min) but i didnt so i swapped around the values and changed the mu and its like we never get back any value but get a new one that's close to what we sub in for C0

basically when i tried to solve μ from equation 3 eg. (ln1.474 / 0.3688) / 0.0025 = 420.8 m-1 which is wrong
 
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