1. Nov 11, 2016

### Hypro96

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 im 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 .

Last edited: Nov 11, 2016
2. Nov 12, 2016

### Hypro96

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 thats 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