Question

What is the mass of 60Co^{60} \mathrm{Co} in a cancer therapy transillumination unit containing 5.00 kCi of 60Co?^{60} \mathrm{Co} ?

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5Here we calculate what is the mass of 60Co{ }^{60} \mathrm{Co} used in cancer therapy. The 60Co{ }^{60} \mathrm{Co} has the activity of 5kCI and consulting the Appendix in the book we can see it has a half-life of 5.271 years, which is 1.66×1081.66 \times 10^{8} s. So, to calculate the mass of this element, firstly we need to remember the definition of activity:

R=0.693Nt1/2R=\frac{0.693 N}{t_{1 / 2}}

where NN is exactly the number of the atoms which will enable us to calculate the mass of the element.

First, we should remember the relation of radioactivity units of Curie and Becquerel, which is: 1 Bq=2.70×1011 Ci1 \mathrm{~Bq}=2.70 \times 10^{-11} \ \mathrm{Ci}.

Following this, now we can convert 5kCi5 \mathrm{kCi} to Becquerels, so we have:

5×103 Ci2.70×1011=1.85×1014 Bq\frac{5 \times 10^{3} \ \mathrm{Ci}}{2.70 \times 10^{-11}} = 1.85 \times 10^{14}\ \mathrm{Bq}

Now expressing the number of the atoms from the first equation we got:

N=R(t1/2)0.693N=\frac{R\left(t_{1 / 2}\right)}{0.693}

Putting in all of the numbers:

N=1.85×1014 Bq(1.66×108 s)0.693N=\frac{1.85 \times 10^{14} \ \mathrm{Bq}\left(1.66 \times 10^{8} \ \mathrm{s}\right)}{0.693}

We get the result of:

N=4.44×1022N=4.44 \times 10^{22}

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