A. 82Rb with T1/2=1.25min, like 81Rb is a positron emitter that is useful for he

Question

A. 82Rb with T1/2=1.25min, like 81Rb is a positron emitter that is useful for heart perfusion imaging. It can be produced by the natural decay of 82Sr (T1/2=25 days). What kind of equilibrium do they undergo?

**I know this answer is Secular Equilibrium but I need help with the following questions:

B. Suppose you have a sample of with 100 mCi of 82Sr. When will the activity of 82Rb reach over 99% of the activity of 82Sr? Answer in minutes.

C. How much daughter activity (Question A) will there be after 20 minutes? Answer in mCi.

D. How much daughter activity (Question A) will there be after 20 days? Answer in mCi.

Explanation / Answer

B. given sample 100 mCi of 82Sr

now activity A = -lambda*N

where, N is the amount of substance ( number of atoms) andd lambda = ln(2)/t1/2

so, 100 mCi = 0.1 ci = 0.1*10^10*3.7 Bq = 3.7*10^9 Bq = lambda*No ( initial amount of 82 Sr = No)

also, t1/2 of 82Sr = 25 days = 25*24*60*60 s

so, No = 3.7*10^9*25*24*60*60 / ln(2) = 1.153*10^16 atoms

now after time t

N atoms of 82Sr have converted into 82Rb

Activity of Sr, A' = lambda(No - N)

Activity of Rb, A' = lambda' ( N)

lambda' = ln(2)/1.25*60

from the given condition

0.99*lambda(No - N) = lambda'(N)

0.99*ln(2)(No - N)/25*24*60*60 = ln(2)*N/1.25*60

0.99*1.25(No - N) = 25*24*60*N

1.2375No - 1.2375N = 36,000 N

N = 3.437No*10^-5

so, time taken = t

but

N = Noe^(-lambda'*t)

(1.153*10^16 - 3.437 *10^-5) = 1.153*10^16*e^(-lambda*t)

ln(0.99999999999999850954032957502168) = -ln(2)*t/25

t = 5.3756969379177915506831485833633e-14 days = 7.7410035906016198329837339600432e-11 minutes

C. After 20 minutes

N = Noexp(-lambda*t)

N = Noexp(-ln(2)*20/25*24*60) [ number of daughter nuclei formed]

activity of N nuclei = -ln(2)*N/1.25*60 = ln(2)Noexp(-ln(2)*20/25*24*60)/1.25*60 = 1.065*10^14 Bq = 2879.20317 Ci

D. After 20 days

N = Noexp(-lambda*t)

N = Noexp(-ln(2)*20/25*24) [ number of daughter nuclei formed]

activity of N nuclei = -ln(2)*N/1.25*60 = ln(2)Noexp(-ln(2)*20/25*24)/1.25*60 = 1.065*10^14 Bq = 2814.5256 Ci

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