A source holds two radioactive isotopes, one short lived and one long lived…

A source holds two radioactive isotopes, one short lived and one long lived. Neither is made from the other. An instrument records the total activity of the source for several days. The figure shows the result on a logarithmic scale. At late times the total follows a straight line, because only the long-lived isotope is left. The dashed line is that straight tail, extended back to time zero. The points marked on the dashed line and on the solid curve are read directly from the figure. Take $\ln 2 = 0.693$ . The source is left in a shielded room and nothing is added to it, so every change in activity comes from decay alone. The readings are corrected for the background count.

A semi-logarithmic graph of the activity of a mixed radioactive source in becquerel, on a logarithmic vertical axis from 100 to 10000, against time in hours from 0 to 90. A solid curve of the total activity starts at 8000 Bq at time zero, bends, and later follows a straight line. A dashed straight line, the extension of that tail, is drawn back to time zero, where it meets the axis at 2000 Bq, and it passes 500 Bq at 60 hours. A point on the solid curve at 30 hours is marked 1750 Bq.

What is the half-life of the long-lived isotope?
What was the activity of the short-lived isotope at time zero?
What is the half-life of the short-lived isotope? Use the marked point at $30\ \text{h}$ .
After how many hours are the activities of the two isotopes equal?
How many nuclei of the long-lived isotope were present at time zero?

Remember that the solid curve is the sum of two activities, and read the point at $30\ \text{h}$ with that in mind.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

(a) Long-lived isotope. On the dashed line the activity falls from $2000\ \text{Bq}$ to $500\ \text{Bq}$ in $60\ \text{h}$ . That is $\dfrac{2000}{500} = 4$ , two halvings, so $T_{\text{long}} = \dfrac{60\ \text{h}}{2} = 30.0\ \text{h}$ .

(b) Short-lived isotope at time zero. The total at time zero is $8000\ \text{Bq}$ , and the long-lived part is $2000\ \text{Bq}$ (the dashed line meets the axis there). So the short-lived part is $8000 - 2000 = 6000\ \text{Bq}$ .

(c) Short-lived isotope. At $30\ \text{h}$ the long-lived part has halved once, to $1000\ \text{Bq}$ . So the short-lived part is $1750 - 1000 = 750\ \text{Bq}$ . It fell from $6000\ \text{Bq}$ to $750\ \text{Bq}$ , a factor of $8$ , which is three halvings in $30\ \text{h}$ . So $T_{\text{short}} = \dfrac{30\ \text{h}}{3} = 10.0\ \text{h}$ .

(d) Equal activities. Set $6000 \times 2^{-t/10} = 2000 \times 2^{-t/30}$ . Then $2^{\,t/10 - t/30} = 3$ , so $\dfrac{t}{15} = \log_2 3 = 1.585$ and $t = 23.8\ \text{h}$ . At that time each activity is $1155\ \text{Bq}$ .

(e) Number of nuclei. $A = \lambda N$ and $\lambda = \dfrac{0.693}{T}$ . So $N = \dfrac{A\,T}{0.693} = \dfrac{2000\ \text{s}^{-1} \times 30 \times 3600\ \text{s}}{0.693} = 3.12 \times 10^{8}$ .

Answer: (a) $30.0\ \text{h}$ . (b) $6000\ \text{Bq}$ . (c) $10.0\ \text{h}$ . (d) $23.8\ \text{h}$ . (e) $3.12 \times 10^{8}$ .

Check. At $60\ \text{h}$ the model gives $6000 \times 2^{-6} + 2000 \times 2^{-2} = 593.8\ \text{Bq}$ , close to the $500\ \text{Bq}$ of the dashed line, as it should be, since the short-lived isotope has almost gone.

Watch out. The point $(30\ \text{h}, 1750\ \text{Bq})$ is on the total curve, not on the dashed line. Using it as if it were the short-lived isotope alone gives a wrong half-life. And $8000\ \text{Bq}$ falling to $1750\ \text{Bq}$ is not three halvings of the whole source.

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