Problem 39 - Fundamentals of Physics

Question

The radionuclide \({ }^{56} \mathrm{Mn}\) has a half-life of \(2.58 \mathrm{~h}\) and is produced in a cyclotron by bombarding a manganese target with deuterons. The target contains only the stable manganese isotope \({ }^{55} \mathrm{Mn},\) and the manganese \(-\) deuteron reaction that produces \({ }^{56} \mathrm{Mn}\) is $${ }^{55} \mathrm{Mn}+\mathrm{d} \rightarrow{ }^{56} \mathrm{Mn}+\mathrm{p}$$ If the bombardment lasts much longer than the half-life of \({ }^{56} \mathrm{Mn}\), the activity of the \({ }^{56} \mathrm{Mn}\) produced in the target reaches a final value of \(8.88 \times 10^{10}\) Bq. (a) At what rate is \({ }^{56}\) Mn being produced? (b) How many \({ }^{56}\) Mn nuclei are then in the target? (c) What is their total mass?

Step-by-Step Solution

Answer

Rate of production is \(8.88 \times 10^{10}\) Bq, number of nuclei is \(1.19 \times 10^{15}\), and mass is \(1.10 \times 10^{-7}\) g.

Step 1: Understanding the Problem

The exercise involves analyzing the production of the radionuclide \(^{56} \mathrm{Mn}\), which has a half-life of 2.58 hours, in a cyclotron. The end production activity is given as \(8.88 \times 10^{10}\) Bq. We are asked to calculate the rate of \(^{56} \mathrm{Mn}\) production, the number of \(^{56} \mathrm{Mn}\) nuclei, and their mass.

Step 1: Calculate the Decay Constant

The decay constant \(\lambda\) is related to the half-life \(t_{1/2}\) of \(^{56} \mathrm{Mn}\) by the formula \(\lambda = \frac{\ln(2)}{t_{1/2}}\). Substituting \(t_{1/2} = 2.58\) hours:\[ \lambda = \frac{\ln(2)}{2.58 \times 3600} = 7.47 \times 10^{-5} \text{ s}^{-1} \]

Step 2: Calculate Rate of Production

The rate of production \(R\) is equal to the activity \(A\) at saturation, where production equals decay. Given the final activity \(A = 8.88 \times 10^{10}\) Bq:\[ R = A = \lambda N = 8.88 \times 10^{10} \text{ Bq}\] So the rate of production is \(8.88 \times 10^{10}\) Bq.

Step 3: Calculate Number of \(^{56} \mathrm{Mn}\) Nuclei

Using the relation \(A = \lambda N\), where \(N\) is the number of nuclei:\[ N = \frac{A}{\lambda} = \frac{8.88 \times 10^{10}}{7.47 \times 10^{-5}} \] \[ N \approx 1.19 \times 10^{15} \text{ nuclei} \]

Step 4: Calculate Total Mass of \(^{56} \mathrm{Mn}\)

First, calculate the mass of a single \(^{56} \mathrm{Mn}\) nucleus using its molar mass \(M = 55.935 \text{ g/mol}\) and Avogadro's number \(N_A = 6.022 \times 10^{23}\) mol\(^{-1}\):\[ \text{mass of one nucleus} = \frac{55.935}{6.022 \times 10^{23}} \text{ g} \] The total mass \(m\) is then:\[ m = N \times \text{mass of one nucleus} \approx 1.19 \times 10^{15} \times \frac{55.935}{6.022 \times 10^{23}} \] \[ m \approx 1.10 \times 10^{-7} \text{ g} \]

Key Concepts

Radioactive Decay

Radioactive decay is a natural process where unstable atomic nuclei lose energy by emitting radiation. This happens as the nucleus seeks a more stable form. During this process, different types of particles can be released, such as alpha particles, beta particles, or gamma rays.
- The decay of a radioactive nucleus can be characterized by its decay constant (\(\lambda\)), indicating the likelihood of decay occurring in a given time period.
- The decay constant is vital in calculating important parameters like the half-life of a substance.
- Radioactive decay follows an exponential law, meaning that the activity of a radioactive substance decreases over time at a rate proportional to its current activity.
In our exercise case, radioactive decay of \(^{56} \mathrm{Mn}\) provides essential parameters like activity and the number of nuclei involved to solve the problem successfully.

Half-Life

Half-life refers to the time required for half the quantity of a radioactive substance to decay. It is a crucial measure in nuclear physics, allowing us to understand how quickly a substance will lose its radioactivity.
- The half-life is inversely related to the decay constant (\(\lambda\)) by the formula \(\lambda = \frac{\ln(2)}{t_{1/2}}\).
- Knowing the half-life helps in determining how often measurements must be taken to manage or harness the material's properties correctly.
- In a practical example like the one given in the step-by-step solution, the half-life of \(^{56} \mathrm{Mn}\) is 2.58 hours, aiding the calculation of its decay constant and further analysis of its production and mass.
These calculations are fundamental in understanding the behavior of radioactive elements and their application in real-world scenarios.

Cyclotron

A cyclotron is a type of particle accelerator used in nuclear physics and chemistry to accelerate charged particles to high energies using a magnetic field and a rapidly varying electric field.
- These high-energy particles are used to bombard targets, such as the manganese in our exercise, initiating nuclear reactions that produce new isotopes.
- In our scenario, the cyclotron enables deuterons to collide with \(^{55} \mathrm{Mn}\) atoms, resulting in the formation of \(^{56} \mathrm{Mn}\).
- Cyclotrons are widely used in medicine, industry, and research for their ability to produce specific isotopes quickly and efficiently.
Understanding the role of a cyclotron in the creation of isotopes is crucial for solving problems related to nuclear reactions and the production of radioisotopes, as illustrated in the exercise provided.