Problem 53
Question
Determining the Size of the Nucless. When radium-226 decays radioactively, it emits an alpha particle (the nucleus of helium), and the end product is radon-222. We can model this decay by thinking of the radium-226 as consisting of an alpha particle emitted from the surface of the spherically symmetric radon-222 nucleus, and we can treat the alpha particle as a point charge. The energy of the alpha particle has been measured in the laboratory and has been found to be 4.79 MeV when the alpha particle is essentially infinitely far from the nucleus. Since radon is much heavier than the alpha particle, we can assume that there is no appreciable recoil of the radon after the decay. The radon nucleus contains 86 protons, while the alpha particle has 2 protons and the radium nucleus has 88 protons. (a) What was the electric potential energy of the alpha-radon combination just before the decay, in MeV and in joules? (b) Use your result from part (a) to calculate the radius of the radon nucleus.
Step-by-Step Solution
Verified Answer
The radius of the radon nucleus is approximately \(7.11 \times 10^{-15}\) meters.
1Step 1: Understand the System
To find the radius of the radon nucleus, we start by visualizing the initial setup. The system consists of an alpha particle and a radon nucleus. The alpha particle can be treated as a point charge moving away from the surface of a spherically symmetric radon-222 nucleus. We know that the radon-222 nucleus has 86 protons, and the alpha particle has 2 protons.
2Step 2: Calculate Initial Electric Potential Energy
The potential energy of the alpha particle when it is close to the radon nucleus is given as total kinetic energy at infinite separation. Using the formula for potential energy between two point charges, \(U = \frac{k \cdot q_1 \cdot q_2}{r}\), where \(k=8.99 \times 10^{9} \, \text{N m}^2/\text{C}^2\,\), the charge of a proton \(e = 1.6 \times 10^{-19} \, \text{C}\), and converting MeV to Joules \(1 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J}\). Alpha's potential energy is thus 4.79 MeV or \(4.79 \times 1.6 \times 10^{-13} \, \text{J} = 7.664 \times 10^{-13} \, \text{J}\).
3Step 3: Convert Potential Energy to Joules
Ensure the initial potential energy calculated is precise. Calculate the electric potential energy in joules as previously outlined. The energy is \( U = 7.664 \times 10^{-13} \, \text{J} \).
4Step 4: Use Potential Energy to Find Radius
Rearranging the potential energy formula, \( r = \frac{k \cdot q_1 \cdot q_2}{U} \), we substitute the values \( q_1 = 2e \) and \( q_2 = 86e \). Substituting the values, we find \( r = \frac{8.99 \times 10^{9} \times 2 \times 86 \times (1.6 \times 10^{-19})^2}{7.664 \times 10^{-13}} \). When solving, ensure all units cancel properly to leave meters.
5Step 5: Solve for Radius
Perform the arithmetic calculations to find the radius: \( r = \frac{8.99 \times 10^{9} \times 2 \times 86 \times (1.6 \times 10^{-19})^2}{7.664 \times 10^{-13}} = 7.11 \times 10^{-15} \, \text{m}\). This result gives us the radius of the radon nucleus.
Key Concepts
Alpha ParticleElectric Potential EnergyRadon Nucleus
Alpha Particle
Understanding an alpha particle is key when studying nuclear decay and reactions. An alpha particle is essentially the nucleus of a helium-4 atom. It consists of two protons and two neutrons, leading to a double positive charge. The significance of the alpha particle lies in its role in alpha decay, a process where a radioactive nucleus emits an alpha particle. Not only does this reduce the atomic number of the original nucleus by two, but it also decreases its mass number by four.
This emission is a common form of radioactive decay among heavy elements like uranium and radium. An alpha particle, due to its relatively large mass and its charge, is not as penetrating as other forms of radiation such as beta particles or gamma rays. Instead, it tends to be stopped by just a few centimeters of air or a sheet of paper, making it less of a concern for shielding compared to other radiation types. Understanding the nature of alpha particles helps in exploring nuclear reactions and the changes in elements after radioactive decay.
This emission is a common form of radioactive decay among heavy elements like uranium and radium. An alpha particle, due to its relatively large mass and its charge, is not as penetrating as other forms of radiation such as beta particles or gamma rays. Instead, it tends to be stopped by just a few centimeters of air or a sheet of paper, making it less of a concern for shielding compared to other radiation types. Understanding the nature of alpha particles helps in exploring nuclear reactions and the changes in elements after radioactive decay.
Electric Potential Energy
Electric potential energy is a form of potential energy associated with the position of charged particles in an electric field. For two point charges, like the alpha particle and the radon nucleus in the decay process, this energy is calculated using the formula:
\[ U = \frac{k \cdot q_1 \cdot q_2}{r} \]
where \( U \) is the electric potential energy, \( k \) is Coulomb's constant \(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\), \( q_1 \) and \( q_2 \) are the charges involved, and \( r \) is the separation distance.
The importance of electric potential energy in nuclear physics is apparent when considering the initial decay scenarios. It dictates how much energy is available to the alpha particle as it moves away from the nucleus. In the given problem, the electric potential energy is initially equivalent to the kinetic energy of the alpha particle when it is infinitely far from the radon nucleus. This equivalence is crucial to understanding not just the energy balance in nuclear reactions but also in calculating further nuclear properties like the radius of the nucleus involved.
\[ U = \frac{k \cdot q_1 \cdot q_2}{r} \]
where \( U \) is the electric potential energy, \( k \) is Coulomb's constant \(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\), \( q_1 \) and \( q_2 \) are the charges involved, and \( r \) is the separation distance.
The importance of electric potential energy in nuclear physics is apparent when considering the initial decay scenarios. It dictates how much energy is available to the alpha particle as it moves away from the nucleus. In the given problem, the electric potential energy is initially equivalent to the kinetic energy of the alpha particle when it is infinitely far from the radon nucleus. This equivalence is crucial to understanding not just the energy balance in nuclear reactions but also in calculating further nuclear properties like the radius of the nucleus involved.
Radon Nucleus
The radon nucleus plays a central role in the decay process mentioned in the problem. Radon-222 refers to an isotope of radon produced from the alpha decay of radium-226. This radon nucleus is vital for understanding the decay chain and its implications. Radon, itself, is a heavy noble gas, known to be radioactive and occurs naturally through the decay of other heavier elements.
The presence of 86 protons in its nucleus characterizes radon-222, reflecting its position on the periodic table and its chemical properties. During the decay process, this nucleus loses an alpha particle, leading to a slight change in structure but maintaining its identity as radon-222 because the number of protons remains the key defining feature.
The presence of 86 protons in its nucleus characterizes radon-222, reflecting its position on the periodic table and its chemical properties. During the decay process, this nucleus loses an alpha particle, leading to a slight change in structure but maintaining its identity as radon-222 because the number of protons remains the key defining feature.
- Understanding the radon nucleus’s role in radioactive decay helps in grasping the energy transformations involved.
- It further aids in nuclear calculations - instances like estimating the radius of the nucleus, based on the potential energy of emitted particles.
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