Question

An \(\alpha\) -particle of energy \(5 \mathrm{MeV}\) is scattered though \(180^{\circ}\) by a fixed uranium nucleus. The distance of the closet approach is of the order of (A) \(10^{-8} \mathrm{~cm}\) (B) \(10^{-12} \mathrm{~cm}\) (C) \(10^{-10} \mathrm{~cm}\) (D) \(10^{-15} \mathrm{~cm}\)

Short answer

The distance of closest approach is of the order of \(10^{-12} cm\). So the answer is (B).

Step-by-step solution

Identify the given information

We are given the following information: - The energy of the alpha particle, E = 5 MeV. We need to convert this to Joules: \(E = 5 \times 10^6 eV \times 1.6 \times 10^{-19} J/eV = 8 \times 10^{-13} J\). - The scattering angle is 180 degrees. - The charge of the alpha particle is 2e, where e is the electron charge, and the charge of the uranium nucleus is 92e (since uranium has an atomic number of 92).

Calculate the electrical potential energy

First, we need to find the electrical potential energy, U, between the alpha particle and the uranium nucleus at the distance of closest approach, d. The formula for electrical potential energy is: \(U = \dfrac{kq_1q_2}{d}\) Where: - k is the Coulomb constant, \(k = 8.99 \times 10^9 Nm^2/C^2\) - \(q_1\) is the charge of the alpha particle, \(q_1 = 2e = 3.2 \times 10^{-19} C\) - \(q_2\) is the charge of the uranium nucleus, \(q_2 = 92e = 1.472 \times 10^{-17} C\)

Use conservation of energy

The law of conservation of energy states that the total energy before and after must stay constant. Since the initial energy was purely kinetic, and at the distance of closest approach all kinetic energy is converted to potential energy, we have: Initial kinetic energy = Final potential energy \(E = U\)

Solve for the distance of closest approach

Now that we have the equation E = U (with E and U having been already calculated), we can solve for d. Substitute \(E\) and \(U\) with the energy and electrical potential energy: \(8 \times 10^{-13} J = \dfrac{(8.99 \times 10^9 Nm^2/C^2)(3.2 \times 10^{-19} C)(1.472 \times 10^{-17} C)}{d}\) Now, isolate "d": \(d = \dfrac{(8.99 \times 10^9 Nm^2/C^2)(3.2 \times 10^{-19} C)(1.472 \times 10^{-17} C)}{8 \times 10^{-13} J}\) And finally, calculate the value for the distance of closest approach: \(d \approx 1.7 \times 10^{-14} m\) Converting this to centimeters: \(d \approx 1.7 \times 10^{-12} cm\) Checking the available options, the distance of the closest approach is of the order of \(10^{-12} cm\). So the answer is (B).

Key concepts

Alpha Particle Scattering

Alpha particle scattering is a fascinating phenomenon in physics that refers to the deflection of alpha particles as they encounter atoms, particularly the nuclei. Imagine a stream of alpha particles directed toward a thin sheet of gold or another heavy nucleus like uranium. Most of these particles pass through with minimal deflection, but a few of them, especially those heading straight for a nucleus, are deflected at various angles, sometimes even bouncing back toward the source. This scattering occurs due to the electrostatic interaction between the positively charged alpha particles and the positively charged nuclei.
- Alpha particles are composed of 2 protons and 2 neutrons, giving them a net charge of +2e.
- The phenomenon was famously used by Ernest Rutherford in 1909, leading to the discovery of the nucleus.
Understanding alpha particle scattering is crucial as it provides insights into atomic structure, particularly about the concentration of positive charge in an atom's nucleus. When an alpha particle travels directly toward a nucleus, it experiences a strong repulsive force due to Coulomb's law, which states that like charges repel.
By observing how alpha particles scatter, scientists were able to conclude that an atom's mass and positive charge are concentrated in a small region called the nucleus, resulting in the planetary model of the atom.

Conservation of Energy

The conservation of energy is a fundamental principle in physics. It states that the total energy of an isolated system remains constant. In the context of alpha particle scattering, this principle ensures that the energy of the alpha particle before and after the collision remains the same, although it may change form.
- Initially, the alpha particle approaches the uranium nucleus with kinetic energy due to its motion.
- As it gets closer and experiences repulsive forces from the nucleus, this kinetic energy is gradually converted into potential energy.

At the point of closest approach, the alpha particle's speed momentarily reaches zero, and all its kinetic energy has been transformed into electric potential energy. This is described mathematically as:\[E_{initial} = U_{final}\]Where:- \(E_{initial}\) is the initial kinetic energy of the alpha particle.- \(U_{final}\) is the potential energy at the closest approach.

This principle helps us determine distances in scenarios like finding the closest approach of an alpha particle to a uranium nucleus. Showing that not only energy conservation laws are universal, but also how they can be practical in solving real-world physics problems.

Potential Energy Calculation

Potential energy calculation during alpha particle scattering is pivotal to understanding the interaction between charged particles. This potential energy is derived from Coulomb's law, which governs the electrical force between two charged objects.
Coulomb's law is expressed as:
\[U = \dfrac{k q_1 q_2}{d}\]Here:
- \(U\) is the electrical potential energy.
- \(k\) is the Coulomb's constant, approximately \(8.99 \times 10^9 \text{Nm}^2/ ext{C}^2\).
- \(q_1\) and \(q_2\) are the charges of the alpha particle and the nucleus, respectively.
- \(d\) is the distance between the charges.

In the scenario of alpha particles approaching a uranium nucleus, the potential energy calculations allow us to determine how close the particle can get before all its kinetic energy is converted into potential energy. By calculating this potential energy at the closest approach, we can effectively determine quantities such as distance, the extent of scattering, and more.

Understanding potential energy calculations is crucial for exploring forces at the microscopic level, such as in nuclear physics, where accurate predictions of particle behavior are necessary for deeper insights into atomic structure and reactions.