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Alejandro Golob
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In this homework question we are told to calculate the half-life of an isotope based on count-rates before and after a given time interval, the relevant equation is given below.
Half-life = t1/2 = tln(2)/ln (A/A0)
The second part asks to determine the standard deviation of the half-life due to counting statistics. We have an uncertainty for the value A and the value A0.
I have found the following solution but cannot quite follow it, so was hoping someone might be able to explain/walk through it with me.
d/dt (t1/2) = − (ln (2) t/(A/A0))/ln((A/A0)2)
Plugging in t = 24 and R = 2.875 yields 5.248 hr−1
σ2t1/2 = (5.248)2 ((9.14/(41.4)2 + 16.83/(118.3)2) *(2.875)2
Thus, we find that expected standard deviation of half-life is σ(t1/2) = 1.22 h
Thanks in advance for any help.
Half-life = t1/2 = tln(2)/ln (A/A0)
The second part asks to determine the standard deviation of the half-life due to counting statistics. We have an uncertainty for the value A and the value A0.
I have found the following solution but cannot quite follow it, so was hoping someone might be able to explain/walk through it with me.
d/dt (t1/2) = − (ln (2) t/(A/A0))/ln((A/A0)2)
Plugging in t = 24 and R = 2.875 yields 5.248 hr−1
σ2t1/2 = (5.248)2 ((9.14/(41.4)2 + 16.83/(118.3)2) *(2.875)2
Thus, we find that expected standard deviation of half-life is σ(t1/2) = 1.22 h
Thanks in advance for any help.