A. Assuming you start with 1 mCi of 52Fe and no 52mMn, what will the activity of
ID: 2078685 • Letter: A
Question
A. Assuming you start with 1 mCi of 52Fe and no 52mMn, what will the activity of 52mMn be after 10 min? Answer in mCi.
B. What will be the activity in previous question (A) after 20 minutes? Answer in mCi.
C. When will the activity of the sample be the highest, i.e. when is the best time to elute the sample? Answer in minutes.
D. Over the weekend there is no one working to elute the sample. Assuming you start with 10Bq of 52Fe and no 52mMn, what will the activity of 52mMn be after 48 hours? Answer in Bq.
E. What is the activity of the 52Fe in Question D at this point? Answe in Bq.
Explanation / Answer
given
1 mCi 52 Fe = 3.7*10^7 Bq
no 52mMn
after 10 minutes
N atoms of Fe have disintegrated
half life of 52Fe, t1/2 = 8.725 hr = 8.725*60*60 s
3.7*10^7 = No*ln(2)/8.725*60*60
No = 1.67665*10^12 atoms of Fe initially
after t = 10 minutes = 10*60 s
atoms of Fe left, N = 1.67*10^12*exp(-10*60*ln(2)/8.725*60*60) = 1.648*10^12 atoms
no of Mn atoms formed 1.67*10^12 - 1.648*10^12 = 2.196*10^10 atoms
half life of Mn = 5.591 day
so activity of Mn = 2.196*10^10*ln(2)/5.591*24*60*60 = 31.510 mBq
2. after 20 minutes
after t = 20 minutes = 20*60 s
atoms of Fe left, N = 1.67*10^12*exp(-20*60*ln(2)/8.725*60*60) = 1.626*10^12 atoms
no of Mn atoms formed 1.67*10^12 - 1.626*10^12 = 4.4*10^10 atoms
half life of Mn = 5.591 day
so activity of Mn = 4.4*10^10*ln(2)/5.591*24*60*60 = 63.135 mBq
3. after time T, activity of sample, for initial No atoms of Fe, halflife of Fe = T and that of Mn be T'
A = ln(2)*No*exp(-ln(2)*t/T)/T + ln(2)*(No(1 - exp(-ln(2)*t/T)))/T'
dA/dt = -ln(2)*ln(2)*No*exp(-ln(2)*t/T)/T*T -ln(2)* ln(2)*(No(1 - exp(-ln(2)*t/T)))/T'*T = 0
-ln(2)*ln(2)*No*exp(-ln(2)*t/T)/T*T =ln(2)* ln(2)*(No(1 - exp(-ln(2)*t/T)))/T'*T
t = Tln((T - T')/T)/ln(2) = 8.725ln((8.725 - 5.591*24)/8.725)/ln(2) = 33.555 hours
4. 10 Bq Fe initially
10 = No*ln(2)/8.725*60*60
after 10 minutes, N = Noexp(-ln(2)*10/8.725*60)
So Mn atoms = No - N
Activity of Mn = (No - N)*ln(2)/5.591*60*60*24 = 8.552*10^-3 Bq
5. Activity of Fe = (Noexp(-ln(2)*10/8.725*60))*ln(2)/8.725*60*60 = 9.868 Bq
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