Activity of cobalt-60

Problem

A sample contains $1{,}0\ \text{mg}$ of $^{60}\text{Co}$ (half-life $T_{1/2} = 5{,}27$ years, atomic mass $\approx 60\ \text{u}$). Calculate the number of atoms present and the initial activity in becquerels. Express the result also in curies ($1\ \text{Ci} = 3{,}7\cdot 10^{10}\ \text{Bq}$).

Solution

Number of atoms. From the mass and the molar mass, via Avogadro’s number $N_A = 6{,}022\cdot 10^{23}$:

$N = \frac{m}{M}\,N_A = \frac{1{,}0\cdot 10^{-3}\ \text{g}}{60\ \text{g/mol}}\cdot 6{,}022\cdot 10^{23} \approx \ev{N \approx 1{,}0\cdot 10^{19}\ \text{atoms}}$

Decay constant. Converting the half-life to seconds ($1\ \text{year} \approx 3{,}156\cdot 10^{7}\ \text{s}$):

$\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0{,}693}{5{,}27\cdot 3{,}156\cdot 10^{7}\ \text{s}} = \frac{0{,}693}{1{,}663\cdot 10^{8}\ \text{s}} \approx 4{,}17\cdot 10^{-9}\ \text{s}^{-1}$

Initial activity.

$A = \lambda N = 4{,}17\cdot 10^{-9}\cdot 1{,}0\cdot 10^{19} \approx \ev{A \approx 4{,}2\cdot 10^{10}\ \text{Bq}}$

In curies:

$A = \frac{4{,}2\cdot 10^{10}}{3{,}7\cdot 10^{10}} \approx \ev{A \approx 1{,}13\ \text{Ci}}$

Typo in the textbook answer key $A \approx 2{,}5\cdot 10^{9}\ \text{Bq} \approx 67\ \text{mCi}$, a value that is wrong and inconsistent with problem 34 of the same book, where $0{,}50\ \text{g}$ of $^{60}\text{Co}$ gives $2{,}09\cdot 10^{13}\ \text{Bq}$. Since activity scales linearly with mass, for $1\ \text{mg}$ it must give $\sim 4{,}2\cdot 10^{10}\ \text{Bq}$. The correct value is adopted here.

The textbook’s answer key gives

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Topics: Nuclear physics Concepts: Radioactive decay