Question 21.SIE.1: Potassium ion is present in foods and is an essential nutrie...
Potassium ion is present in foods and is an essential nutrient in the human body. One of the naturally occurring isotopes of potassium, potassium-40, is radioactive. Potassium-40 has a natural abundance of 0.0117 % and a half-life t_{1/2} = 1.28 × 10^9 yr. It undergoes radioactive decay in three ways: 98.2 % is by electron capture, 1.35% is by beta emission, and 0.49 % is by positron emission. (a) Why should we expect ^{40}K to be radioactive? (b) Write the nuclear equations for the three modes by which ^{40}K decays. (c) How many ^{40}K^+ ions are present in 1.00 g of KCl? (d) How long does it take for 1.00% of the ^{40}K in a sample to undergo radioactive decay?
(a) The ^{40}K nucleus contains 19 protons and 21 neutrons. There are very few stable nuclei with odd numbers of both protons and neutrons (Section 21.2).
(b) Electron capture is capture of an inner-shell electron by the nucleus:
{}_{19}^{40}K +{}_{-1}^0 e \longrightarrow{}_{18}^{40}Ar
Beta emission is loss of a beta particle \left({}_{-1}^0 e \right) by the nucleus:
{}_{19}^{40}K \longrightarrow{}_{20}^{40}Ca +{}_{-1}^0 e
Positron emission is loss of a positron \left(\begin{array}{r}{}_{+1}^0 e\end{array}\right) by the nucleus:
{}_{19}^{40}K \longrightarrow{}_{18}^{40}Ar +{}_{+1}^0 e
(c) The total number of K^+ ions in the sample is:
(1.00 \,\cancel{g \,KCI} )\left(\frac{1\, \cancel{mol \,KCI}}{74.55\, \cancel{g\, KCI}}\right)\left(\frac{1\, \cancel{mol\, K ^{+}}}{1\, \cancel{mol\, KCI}}\right)\left(\frac{6.022 \times 10^{23}\,K ^{+}}{1 \,\cancel{mol \,K ^{+}}}\right)=8.08 \times 10^{21}\,K ^{+}\text{ions}
Of these, 0.0117% are ^{40}K^+ ions:
\left(8.08 \times 10^{21}\,\cancel{K ^{+}\,ions}\right)\left(\frac{0.0117^{40}K ^{+}\text{ions}}{100 \cancel{K ^{+}\,ions}}\right)=9.45 \times 10^{17}{^{40}K }^{+}\text{ions}
(d) The decay constant (the rate constant) for the radioactive decay can be calculated from the half-life, using Equation 21.19:
k=\frac{0.693}{t_{1 / 2}} (21.19)
k=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{1.28 \times 10^9 yr}=\left(5.41 \times 10^{-10}\right) / yr
The rate equation, Equation 21.20, then allows us to calculate the time required:
\ln \frac{N_t}{N_0}=-k t (21.20)
\begin{aligned}\ln \frac{N_t}{N_0}&=-k t \\\ln \frac{99}{100}&=-\left[\left(5.41 \times 10^{-10}\right) / yr \right] t \\-0.01005 &=-\left[\left(5.41 \times 10^{-10}\right) / yr \right] t \\t &=\frac{-0.01005}{\left(-5.41 \times 10^{-10}\right) / yr}=1.86 \times 10^7 yr\end{aligned}
That is, it would take 18.6 million years for just 1.00% of the ^{40}K in a sample to decay.