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Chapter 21: Problem 90

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. Assume 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 \(\mathrm{J} / \mathrm{s}\).).

Short Answer

Expert verified
The decay equation is \({ }^{238} Pu \rightarrow { }^{234} U + { }^{4} He\). The power output at t=0 and t=10 years would need to be calculated using the methodology outlined in Steps 3 and 4, respectively.

Step by step solution

01

Create Decay Formula

The decay equation for Plutonium-238 will involve the emission of an alpha particle (denoted as \({ }^{4} He\)). So, when plutonium-238 decays, it transforms into Uranium-234 and emits an alpha particle, which would be written as \({ }^{238} Pu \rightarrow { }^{234} U + { }^{4} He\). This is the decay process for plutonium-238.
02

Calculate Decay Constant

The decay constant (λ) is related to the half-life (\(T_{1/2}\)) by the equation \(\lambda = \ln(2) / T_{1/2}\). Given that the half-life of plutonium-238 is 86 years, substituting this into the above equation gives \(\lambda = \ln(2) / 86\).
03

Power Output Calculation at t=0

The power output at t=0 is the power output when all the Pu-238 decays. We can calculate this by determining the number of atoms in 1 mg of Pu-238 and then multiplying this by the energy per decay. \nNumber of atoms can be calculated using Avogadro's number and the molar mass. For Pu-238:\nNumber of atoms = (1mg / 238g mol−1) × Avogadro's number \nThe power output (P) at t=0 would then be calculated as:\nP = Energy per decay × Number of atoms decaying per unit time. Since all the atoms decay at t=0, the time interval considered is one second.
04

Power Output Calculation at t=10 years

Using the decay equation from Step 1, we calculate the remaining activity after 10 years given that the activity decreases by 8%. The remaining activity is \(N = N_{0} × 0.92\) (0.92 because 8% has decayed). We use this remaining activity to calculate the power at t=10 years.

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