The power supply for a pacemaker is a small amount of radioactive $^{238} \mathrm{Pu} .\( This nuclide decays by \)\alpha$ decay with a half-life of 86 yr. The pacemaker is typically replaced every 10.0 yr. (a) By what percentage does the activity of the \(^{238}\) Pu source decrease in 10 yr? (b) The energy of the \(\alpha\) particles emitted is 5.6 MeV. Assume an efficiency of \(100 \%\) -all of the \(\alpha\) -particle energy is used to run the pacemaker. If the pacemaker starts with \(1.0 \mathrm{mg}\) of \(^{238} \mathrm{Pu}\) what is the power output initially and after 10.0 yr?

For radioactive decay, we use the formula \(N(t) = N_0 e^{-\lambda t}\), where \(N(t)\) is the number of radioactive atoms at time t, \(N_0\) is the initial number of radioactive atoms, \(\lambda\) is the decay constant, and t is the time elapsed. First, we need to find the decay constant \(\lambda\) using the half-life formula: \(T_{1/2} = \frac{ln(2)}{\lambda}\). Given the half-life of 86 years, we can solve for the decay constant: $$\lambda = \frac{ln(2)}{86} = 0.00805 \; \text{yr}^{-1}$$

We can find the percentage decrease in activity by first finding the ratio of atoms remaining after 10 years to the initial number of atoms. Using the formula \(N(t) = N_0 e^{-\lambda t}\), we have: $$\frac{N(10)}{N_0} = e^{-\lambda \cdot 10} = e^{-0.00805 \cdot 10} = 0.924$$ Next, we find the percentage decrease by subtracting the ratio from 1 and multiplying by 100: $$\%\text{decrease} = (1 - \frac{N(10)}{N_0}) \cdot 100 = (1 - 0.924) \cdot 100 = 7.6 \%$$

We're given that the energy of the emitted \(\alpha\) particles is 5.6 MeV per decay. Since the total energy output depends on the decay rate, we have to find the number of decays per second from the initial \(1.0 \;\text{mg}\) of \(^{238}\mathrm{Pu}\). Converting mass to number of atoms, we have: $$N_0 = \frac{1.0 \;\text{mg}}{238 \;\text{amu}} \cdot 6.022 \cdot 10^{23} \;\text{atoms/mol} = 2.54 \cdot 10^{21} \;\text{atoms}$$ Now, we can find the activity (decay rate) of the \(^{238}\mathrm{Pu}\) with the relation \(A = \lambda N_0\): $$A_0 = \lambda N_0 = 0.00805 \; \text{yr}^{-1} \cdot 2.54 \cdot 10^{21} \;\text{atoms} = 2.045 \cdot 10^{19} \;\text{decays/year}$$ We need to convert this decay rate to seconds: $$A_0 = 2.045 \cdot 10^{19} \frac{\text{decays}}{\text{year}}\cdot \frac{1 \;\text{year}}{3.1536 \times 10^7 \;\text{seconds}} = 6.48 \cdot 10^{11} \frac{\text{decays}}{\text{second}}$$ Now, we can find the power output by multiplying the energy per decay by the activity: $$P_0 = E \cdot A_0 = (5.6 \;\text{MeV})(6.48 \cdot 10^{11} \frac{\text{decays}}{\text{second}}) = 3.62 \cdot 10^{12} \;\text{MeV/s}$$ We need to convert this to more convenient units, such as watts. Knowing that \(1 \;\text{MeV} = 1.602 \times 10^{-13} \;\text{J}\), we have: $$P_0 = 3.62 \cdot 10^{12} \;\text{MeV/s} \cdot 1.602 \times 10^{-13} \;\text{J/MeV} = 0.58 \;\text{W}$$

We know that the activity decreases by 7.6% after 10 years, so the new activity is: $$A_{10} = A_0 (1 - 0.076) = 6.48 \cdot 10^{11} \frac{\text{decays}}{\text{second}} \cdot (1 - 0.076) = 5.98 \cdot 10^{11} \frac{\text{decays}}{\text{second}}$$ Now, we can find the power output at t=10 years: $$P_{10} = E \cdot A_{10} = (5.6 \;\text{MeV})(5.98 \cdot 10^{11} \frac{\text{decays}}{\text{second}}) = 3.35 \cdot 10^{12} \;\text{MeV/s}$$ Converting to watts, we have: $$P_{10} = 3.35 \cdot 10^{12} \;\text{MeV/s} \cdot 1.602 \times 10^{-13} \;\text{J/MeV} = 0.54 \;\text{W}$$ (a) The activity of the \(^{238}\mathrm{Pu}\) source decreases by 7.6% in 10 years. (b) The initial power output of the pacemaker is 0.58 W, and after 10 years, the power output is 0.54 W.