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Exercises · 13.20

Q.Obtain the amount of 2760Co^{60}_{27}\text{Co} necessary to provide a radioactive source of 8.0 mCi strength. The half-life of 2760Co^{60}_{27}\text{Co} is 5.3 years.

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Convert 8.0 mCi to decays/second, find the decay constant from the half-life, solve A=λNA=\lambda N for NN, then convert NN to mass. The result is a tiny mass, about 7.1 μ\mug — a reminder of how enormously radioactive even a microgram-scale sample of a short-half-life isotope is.

Step 1 — Convert the required activity to SI units

1 Ci=3.7×1010 decays/s1\ \text{Ci} = 3.7 \times 10^{10}\ \text{decays/s}

A=8.0 mCi=8.0×10−3×3.7×1010=2.96×108 decays/sA = 8.0\ \text{mCi} = 8.0\times 10^{-3} \times 3.7\times 10^{10} = 2.96\times 10^{8}\ \text{decays/s}

Step 2 — Decay constant from the half-life

T1/2=5.3 yr=5.3×3.154×107 s=1.6716×108 sT_{1/2} = 5.3\ \text{yr} = 5.3 \times 3.154\times 10^{7}\ \text{s} = 1.6716\times 10^{8}\ \text{s}

λ=ln⁡2T1/2=0.6931471.6716×108=4.147×10−9 s−1\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693147}{1.6716\times 10^{8}} = 4.147\times 10^{-9}\ \text{s}^{-1}

Step 3 — Number of atoms needed

Activity A=λNA = \lambda N, so: …

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