Question 42.36: Plutonium isotope ^239Pu decays by alpha decay with a half-l......
Plutonium isotope ^{239}Pu decays by alpha decay with a half-life of 24 100 y. How many milligrams of helium are produced by an initially pure 12.0 g sample of ^{239}Pu at the end of 20 000 y? (Consider only the helium produced directly by the plutonium and not by any by-products of the decay process.)
We have one alpha particle (helium nucleus) produced for every plutonium nucleus that decays. To find the number that have decayed, we use Eq. 42-15, Eq. 42-18, and adapt Eq. 42-21:
N=N_0 e^{-\lambda t} \quad \text { (radioactive decay), } (42-15)
T_{1 / 2}=\frac{\ln 2}{\lambda}=\tau \ln 2 . (42-18)
N_{40}=\left(1.17 \times 10^{-4}\right) \frac{M_{ sam } N_{ A }}{M} (42-21)
N_0-N=N_0\left(1-e^{-t \ln 2 / T_{1 / 2}}\right)=N_A \frac{12.0 \mathrm{~g} / \mathrm{mol}}{239 \mathrm{~g} / \mathrm{mol}}\left(1-e^{-20000 \ln 2 / 24100}\right)
where N_A is the Avogadro constant. This yields 1.32 \times 10^{22} alpha particles produced. In terms of the amount of helium gas produced (assuming the α particles slow down and capture the appropriate number of electrons), this corresponds to
m_{\mathrm{He}}=\left(\frac{1.32 \times 10^{22}}{6.02 \times 10^{23} / \mathrm{mol}}\right)(4.0 \mathrm{~g} / \mathrm{mol})=87.9 \times 10^{-3} \mathrm{~g} .