The radioactive isotope \({ }^{238} \mathrm{Pu},\) used in pacemakers, decays by emitting an alpha particle with a half-life of 86 yr. (a) Write an equation for the decay process. (b) The energy of the emitted alpha particle is \(9.0 \times 10^{-13} \mathrm{~J},\) which is the energy per decay. Assuming that all the alpha particle energy is used to run the pacemaker, calculate the power output at \(t=0\) and \(t=10 \mathrm{yr} .\) Initially \(1.0 \mathrm{mg}\) of \({ }^{238} \mathrm{Pu}\) was present in the pacemaker. (Hint: After \(10 \mathrm{yr}\), the activity of the isotope decreases by 8.0 percent. Power is measured in watts or \(\mathrm{J} / \mathrm{s}\).

In alpha decay, the parent nucleus loses two protons and two neutrons to form a daughter nucleus and an alpha particle (a helium nucleus, He). So for Pu-238:\[{}^{238}_{\ 94}Pu \rightarrow {}^{234}_{\ 92}U + {}^{4}_{\ 2}He\]Note that the subscript is the atomic number (number of protons) and the superscript is the atomic weight (number of protons + neutrons). Both sides of the equation add up correctly.

We know the half-life of Pu-238 is 86 years. The decay constant \(\lambda\) is related to this by the following equation:\[\lambda = \frac{ln(2)}{T_{1/2}}\]Where \(T_{1/2}\) = 86 years = \(2.711 \times 10^9\) seconds. So \(\lambda \approx 2.552 \times 10^{-17}s^{-1}\).The initial number of nuclei N(0) in 1 mg of Pu-238 can be calculated using Avogadro's number and the atomic mass of Pu-238:\[N(0) = \frac{(1mg)(6.022 \times 10^{23} atoms/mol)}{238 g/mol} \approx 2.532 \times 10^{18} atoms\]So the initial activity A(0) = \(\lambda N(0)\) is approximately\[A(0) = \lambda N(0) \approx 2.552 \times 10^{-17}s^{-1} \times 2.532 \times 10^{18} atoms \approx 64.62 decay/s\]This is how many atoms decay per second at the start.

Each decay gives 9.0 × 10^{-13} J of energy, so the initial power is \(P(0) = A(0)E\), where E is the energy per decay:\[P(0) = A(0)E = 64.62 decay/s \times 9.0 × 10^{-13}J/decay = 5.82 \times 10^{-11} W\]This is the power output at the start (t=0).

After 10 years, the activity has decreased by 8.0 %. So at t=10 yr, the activity is 92% of its initial value:\[A(10) = 0.92 A(0) = 0.92 \times 64.62 decay/s = 59.45 decay/s\]So the power at t=10 yr is \[P(10) = A(10)E = 59.45 decay/s \times 9.0 × 10^{-13}J/decay = 5.35 \times 10^{-11} W\]This is the power output after 10 years.