Question

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

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

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.

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 \(\lambda = \ln(2) / 86\).

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.

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.

Key concepts

Alpha Particle Emission

In radioactive decay, alpha particle emission is a common process where an unstable nucleus sheds energy by releasing an alpha particle, which consists of two protons and two neutrons (essentially a helium nucleus: \( ^{4} He \)).
Alpha emission results in the transformation of the decaying atom into a new element with a lower atomic number by two and mass number by four. For example, in the decay of Plutonium-238 \(( ^{238} Pu )\), it emits an alpha particle and transforms into Uranium-234 \(( ^{234} U )\).
This is written as the decay equation:

• \( ^{238} Pu \rightarrow ^{234} U + ^{4} He \)

This type of decay is indicative of the heavier elements found at the end of the periodic table and is an essential concept in understanding radioactive behaviors.

Half-Life

The half-life of a radioactive isotope is the time taken for half of the nuclei in a sample to undergo decay. It helps to predict how long an isotope remains active or useful.
Plutonium-238 has a half-life of 86 years, meaning, every 86 years, the amount of original Pu-238 is halved.
Knowing the half-life allows us to calculate how much of the substance remains at any given time and is crucial for evaluating the longevity and power potential of radioactive materials.
Using the formula:

• Decay constant \( \lambda = \ln(2) / T_{1/2} \), where \( T_{1/2} \) is the half-life in years.

Understanding half-life is vital for applications like pacemakers, where longevity and consistency in power output are necessary.

Power Calculation

Power calculation in the context of radioactive decay involves determining the energy output over time as a result of isotope decay.
The power output can be calculated by understanding the number of decays per unit time and the energy released per decay.
In the case of Plutonium-238:

• The number of atoms initially present in 1 mg can be found using Avogadro's number and the molar mass (238 g/mol).
• Power (\( P \)) is computed as \( P = \text{Energy per decay} \times \text{Number of atoms decaying per unit time} \).

At \( t=0 \), all available Pu-238 atoms are considered to decay, providing maximum initial power, and this decreases over time as the isotope decays further.

Isotopic Decay

Isotopic decay refers to the transformation of an unstable isotope into a more stable form or a different element altogether. This happens through the emission of particles, in this case, alpha particles.
Each decay reduces the number of initial atoms, leading eventually to a stable substance or another isotope.
Plutonium-238, for example, decays into Uranium-234 by emitting an alpha particle, contributing to energy changes within a material.
Understanding isotopic decay helps in predicting the lifecycle of radioactive substances and aligns with calculations of residual activities over time.

Energy Per Decay

Energy per decay quantifies the energy released during each decay event of a radioactive nucleus. For Plutonium-238, the energy per decay due to alpha emission is \( 9.0 \times 10^{-13} \text{ J} \).
This energy is critical for applications where every decay contributes directly to the device's operation, such as the continuous powering of a pacemaker.

• Energy measurements allow for assessing the initial and subsequent power outputs over time.
• The calculated energy per decay helps in predicting device lifetime and efficiency.

Being efficient in converting decays to energy output is a key factor in device design and functionality.