Question

A 4.50-MeV alpha particle is incident on a platinum nucleus \((Z=78)\). What is the minimum distance of approach, \(r_{\min } ?\)

Short answer

Answer: The minimum distance of approach of the alpha particle to the platinum nucleus is approximately \(2.91\times10^{-14}\,\text{m}\).

Step-by-step solution

Find the electrostatic force between the alpha particle and the platinum nucleus

The electrostatic force between two charged particles can be calculated using the Coulomb's law: $$F = \frac{k\cdot{}q_1\cdot{}q_2}{r^2}$$ where F is the force, k is the electrostatic constant (approximately \(8.99\times10^9\,\text{N m}²/\text{C}²\)), \(q_1\) and \(q_2\) are the charges of the particles, and r is the distance between them. For an alpha particle, the charge is \(q_1=+2e\) where e is the elementary charge (\(1.6\times10^{-19}\,\text{C}\)). For a platinum nucleus with Z=78, the charge is \(q_2=+78e\).

Apply the conservation of energy

According to the conservation of energy, the total mechanical energy (the sum of kinetic energy and potential energy) of a system of particles remains constant unless some external force acts on it. In this case, the electrostatic force is an internal force, so we can apply the conservation of energy. Initially, the alpha particle has a kinetic energy of 4.50 MeV, and its potential energy is given by $$U = \frac{k\cdot{}q_1\cdot{}q_2}{r}$$. At the minimum distance of approach, the kinetic energy of the alpha particle will be zero, so all its energy will be potential energy. The energy condition can be written as follows: $$4.50\,\text{MeV} = \frac{k\cdot{}q_1\cdot{}q_2}{r_\mathrm{min}}$$

Convert the energy to the same unit and solve for \(r_\mathrm{min}\)

The given energy is in MeV, so we need to convert it to joules using the conversion factor: 1 MeV = \(1.6\times10^{-13}\,\text{J}\). So, the initial kinetic energy of the alpha particle in joules is $$4.50\,\text{MeV} \cdot (1.6\times10^{-13}\,\text{J/MeV}) = 7.2\times10^{-13}\,\text{J}$$. Now, plug the values of k, \(q_1\), \(q_2\), and the energy into the equation derived in step 2 and solve for \(r_\mathrm{min}\): $$7.2\times10^{-13}\,\text{J}=\frac{ (8.99\times10^9\,\text{N m}²/\text{C}²)(2e)(78e)}{r_\mathrm{min}}$$ $$r_\mathrm{min}=\frac{(8.99\times10^9\,\text{N m}²/\text{C}²)(2\times1.6\times10^{-19}\,\text{C})(78\times1.6\times10^{-19}\,\text{C})}{7.2\times10^{-13}\,\text{J}}=$$ $$\approx 2.91\times10^{-14}\,\text{m}$$ Therefore, the minimum distance of approach of the alpha particle to the platinum nucleus is approximately \(2.91\times10^{-14}\,\text{m}\).

Key concepts

Coulomb's Law

Understanding the intricate dance of charged particles starts with Coulomb's law, which provides the mathematical expression for the electrostatic force between two point charges. The equation, represented as \[ F = \frac{k\cdot q_1\cdot q_2}{r^2} \], encapsulates how the force (F) between charges is directly proportional to the product of the charges themselves (\(q_1 \text{ and } q_2\) and inversely proportional to the square of the distance (r) between them.

The constant k, known as Coulomb's constant, has an approximate value of \(8.99\times10^9\,\text{N m}^2/\text{C}^2\). Thus, as the alpha particle and the platinum nucleus in our exercise scenario approach each other, the force increases significantly when the distance decreases, which ultimately determines the minimum distance of approach before the repulsive forces counterbalance the kinetic energy of the alpha particle.

Conservation of Energy

The conservation of energy principle is a fundamental concept that asserts the total energy in an isolated system remains constant over time. For our exercise, we focus on how the kinetic energy of a moving alpha particle is converted into electrostatic potential energy as it approaches a platinum nucleus.

Initially, the alpha particle's kinetic energy is given, and the potential energy can be defined using Coulomb's law. The key to solving the problem is to set these two energies equal to each other at the point where the kinetic energy is fully converted into potential energy, indicating the closest approach. No external work is done on the system, thus honoring the principle of conservation of energy.

Kinetic Energy

Kinetic energy is the energy possessed by an object in motion. For charged particles like an alpha particle, this energy can be quantified and is crucial in determining particle interactions. In our problem, the initial kinetic energy of the particle is given and is used to find the minimum distance of approach to the platinum nucleus.

The initial kinetic energy (4.50 MeV for the alpha particle) represents the capability of the particle to do work on its surroundings, in this case by moving against the electrostatic force of the platinum nucleus. As the particle moves closer, this kinetic energy decreases, eventually reaching zero at the minimum distance of approach.

Electrostatic Force

The invisible push and pull between charged particles is characterized by the electrostatic force. It is a vector force meaning it has both magnitude and direction, and is one of the principal forces responsible for the interactions between atomic particles.

In our case, we analyze how the electrostatic repulsion between the positively charged alpha particle and the platinum nucleus reaches a balance point, where the alpha particle can get no closer — the minimum distance of approach. This force becomes formidable as the distance shrinks, clearly laying out the battleground where the initial kinetic energy of the alpha particle is exhaustively converted into potential — a classic display of electrostatic force at play.