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Rectifier

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**The problem statement**

The Cobalt-60 isotope is often used in medicine. It has a half-life of 5.25 years.

a) What is the activity of 1μg Cobalt in Bq (decays/second)?

b) You find a sample of Cobalt that has a mass of 10 µg and has an activity of 5.0MBq. How old is that sample?

*This problem was translated from Swedish.*

**The attempt at a solution**

a)

Activity is given by:

## A_{Bq}= n N_A \frac{ln(2)}{t_{\frac{1}{2}}} ##

where ## t_{\frac{1}{2}} = 1.656 \cdot 10^8 ## since ## 5.25 years =1.656 \cdot 10^8 seconds ## and ##N_A## is avogados constant which is ## 6.022×10^{23} mol^{−1} ## but what is n?

I guess that that is the amount of moles in 1μg Cobalt. Wolfram tells me its

## 1.6968366×10^{-8} mol ##

Thus the activity is

## A_{Bq}= n N_A \frac{ln(2)}{t_{\frac{1}{2}}} = 1.6968366 \cdot 10^{-8}\cdot 6.022\cdot10^{23}\cdot \frac{ln(2)}{1.656 \cdot 10^8}=

4.27707 \cdot 10^7##

b)

## A(t) = A_0 0.5^{\frac{t}{ t_{\frac{1}{2}} }} ##

where

A(t) = 5.0MBq

A(0) is the activity of 1μg Cobalt

##t_{\frac{1}{2}}## = ## 1.656 \cdot 10^8 ##

## A(t) = A_0 0.5^{\frac{t}{ t_{\frac{1}{2}} }} \\ 5.0 \cdot 10^6 = 4.27707 \cdot 10^7 \cdot 0.5^{\frac{t}{ 1.656 \cdot 10^8} } \\ t =5.12801×10^8 s ##

Which is about 16.26 years.

Is that solution correct?

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