Solve the thickness of paper needed to block 99% of Beta rays from Strontium. So
ID: 2076534 • Letter: S
Question
Solve the thickness of paper needed to block 99% of Beta rays from Strontium.Solve the thickness of iron needed to block 99% of Gamma rays. I'm quiet confused on how to solve for these. It also states that when 99% of Beta radiation remains, then only 1% of the radiation intensity remains. Everything you need to solve it should be there, please help? diation composition of different sources: gamma beta gamma beta enetration factors for Brays from Sr passing through pa 17, soo Initial radiation intensity (full emission): CPM Final radiation intensity with 50 absorbing cards: CPM Thickness of 50 absorbing cards in centimeters: Penetration factor for Sr passing through paper: (Show calculations) 60) to 3 you Thickness of paper needed to block 99% of the B rays from sr: cm (Show calculations) assing through iron Initial radiation intensity (with on at CPM CPM Final radiation intensity with 5 absorbing plates: Thickness of 4 absorbing plates in centimeters: cm, Penetration factor for Co passing through iron: (Show calculations) Thickness of iron needed to block 99% ofthe y rays from 60Co: cm (Show calculations)
Explanation / Answer
the intensity due to sheilding of a paper of thickness x of linear attenuation factor n is given by
IB = IA e-nx
where IB and IA are the final and initial intensity respectively.
here IB /IA = 0.01,
thus, e-nx = 0.01
in the case of paper x = 1 cm.
hence 450 = 17500e-nx
here x = 1cm.
n = ln(0.02571)
n= 3.66;
there thickness of the paper for 99% attenuation of beta ray is as below
x = (-1/3.66)*ln(0.01) = 1.258 cm
In the second case for one plate the intensity is 1000 cpm keeping 4 additional iron plates the intensity become 600 CPM. the thickness of the 4 plates are 1.78 cm suppose the attenuation factor is n thus
600 = 1000* exp(-n*1.78);
hence n = 0.399
let for 99% attenuation the thickness of the iron plate is t
0.01 = exp(-0.399 * t)
t = -(1/0.399)*ln(0.01) = 11.542 cm