Problem 51 - University Physics with Modern Physics

Question

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

Answer

(a) 4.79 MeV, or 7.67 x 10^-13 J. (b) The radius of the radon nucleus is approximately 7.65 x 10^-15 m.

Step 1: Understanding Electric Potential Energy

The electric potential energy (U) between two point charges can be calculated using Coulomb's law. The formula is given as:\[ U = \frac{k \cdot q_1 \cdot q_2}{r} \]where \(k\) is Coulomb's constant \( (8.99 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2)\), \(q_1\) and \(q_2\) are the charges of the two particles, and \(r\) is the separation distance, which in this case is the radius of the radon nucleus. Given the conditions of the problem, we'll first solve for \( U \).

Step 2: Calculate the Charge Products

Identify the charges:- The alpha particle has a charge of \(2e\).- The radon nucleus has a charge of \(86e\).Plug these values into the equation to get:\[ q_1 \cdot q_2 = (2e)(86e) = 172e^2 \]where \(e = 1.602 \times 10^{-19} \text{ C}\).

Step 3: Relate Energy Released to Electric Potential Energy

Before the decay, the total energy of the alpha particle is equal to its electric potential energy, which is converted into the kinetic energy of \(4.79\) MeV as the alpha particle moves far away. Hence:\[ U = 4.79 \text{ MeV } \times 1.602 \times 10^{-13} \text{ J/MeV} \approx 7.66758 \times 10^{-13} \text{ J} \].

Step 4: Calculate the Radius of the Radon Nucleus

Using \( U = \frac{k \cdot q_1 \cdot q_2}{r} \), rearrange to solve for \( r \):\[ r = \frac{k \cdot q_1 \cdot q_2}{U} \]Substitute the known values:\[ r = \frac{(8.99 \times 10^9) \times (172\times (1.602 \times 10^{-19})^2)}{7.66758 \times 10^{-13}} \]Calculate \( r \) in meters.

Step 5: Plug in the Values and Solve for the Radius

Perform the calculation for \( r \):\[ r = \frac{(8.99 \times 10^9) \times 172 \times (2.56 \times 10^{-38})}{7.66758 \times 10^{-13}} \]\[ r \approx 7.65 \times 10^{-15} \text{ m} \]This is the radius of the radon nucleus.

Key Concepts

Electric Potential Energy

Electric potential energy is a form of energy that arises from the interaction between electrically charged particles. It can be thought of as the potential energy that a charged particle possesses due to its position relative to other charged particles.

In the context of nuclear physics, electric potential energy plays a crucial role in understanding how charged particles interact within an atomic nucleus. When discussing radioactive decay, such as the alpha decay of radium-226, the electric potential energy between the alpha particle (which has a positive charge) and the radon nucleus (also positively charged) is of particular interest.

Before the decay, the potential energy is stored in the system. After the alpha particle is emitted, this energy is converted, mainly into the kinetic energy of the alpha particle, as seen in the 4.79 MeV energy measurement. Understanding how this energy changes is key to predicting particle behavior.

Coulomb's Law

Coulomb's law is fundamental to understanding the electric forces between charged particles. It describes how the electric force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

The mathematical expression for this law is given as: \[ F = \frac{k \cdot q_1 \cdot q_2}{r^2} \] where \( F \) is the force, \( k \) is Coulomb's constant \((8.99 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2)\), \( q_1 \) and \( q_2 \) are the charges of the particles, and \( r \) is the distance separating them.

In the process of alpha decay, Coulomb's law allows us to compute the electric potential energy of the system before the decay occurs. By knowing the charges of the alpha particle and the radon nucleus, we can use these values in Coulomb's formula to determine the potential energy, which is indicative of the repulsive electric force experienced due to like charges.

This computation helps us estimate the radius of the nucleus, as the distance \( r \) influences both the potential energy and the force experienced by the alpha particle.

Nuclear Physics

Nuclear physics is the field of physics that studies the constituents and interactions of atomic nuclei. It explores the binding forces holding the nucleus together and the processes by which nuclei emit particles such as in radioactive decay.

A key aspect of nuclear physics is its focus on phenomena like alpha decay. In this process, an unstable nucleus releases an alpha particle, reducing its atomic number and transforming into a different element. The emission of an alpha particle is a way for the nucleus to attain a more stable energy state.

The study encompasses understanding energy interactions, particle behavior, and theoretical models that explain these nuclear changes. It has profound implications for applications like nuclear power and medical therapy.

In the case of radium-226 decaying into radon-222, investigating these interactions involves considering factors like electric potential energy and the rules governing nuclear decay, which helps us comprehend the decay's effects on both the particles involved and the energy released.

Radioactive Decay

Radioactive decay is a natural process by which an unstable atomic nucleus loses energy by emitting radiation. This decay can result in the emission of particles like alpha and beta particles, or electromagnetic radiation in the form of gamma rays.

Alpha decay specifically refers to the emission of an alpha particle, which consists of two protons and two neutrons. This process results in the reduction of the atomic number by two units and the mass number by four units, leading to the formation of a new element.

For radium-226, undergoing alpha decay produces radon-222 and an alpha particle. This transformation is not only a result of the need to achieve stability but is also an indicator of the energy exchange within the nucleus. The energy released, as in the 4.79 MeV of kinetic energy, is initially stored as electric potential energy in the nucleus before decay.

Understanding radioactive decay is crucial in fields such as medicine, where radioactive isotopes are used for diagnosis and treatment, and in nuclear energy, where it serves as a power source.