Problem 48 - Master Resource Book in JEE Main Physics

Question

An \(\alpha\)-particle of energy \(5 \mathrm{MeV}\) is scattered through \(180^{\circ}\) by a fixed uranium nucleus. The distance of closest approach is of the order of (a) \(1 \mathrm{~A}\) (b) \(10^{-10} \mathrm{~cm}\) (c) \(10^{-12} \mathrm{~cm}\) (d) \(10^{-15} \mathrm{~cm}\)

Step-by-Step Solution

Answer

The distance of closest approach is about \(10^{-15} \mathrm{~cm}\). Answer: (d).

Step 1: Understand the Problem

The problem involves an \(\alpha\)-particle with a given energy being scattered by a uranium nucleus. We need to determine the distance of closest approach, which is where the kinetic energy of the \(\alpha\)-particle is completely converted to electrostatic potential energy due to Coulomb interaction.

Step 2: Recall the Formula for Closest Approach

The distance of closest approach \(d\) can be found using the formula \[ d = \frac{k_e \cdot Z_1 Z_2 e^2}{2E} \] where \(k_e\) is Coulomb's constant \(8.988 \times 10^9 \mathrm{~N~m^2/C^2}\), \(Z_1 = 2\) is the atomic number of the \(\alpha\)-particle, \(Z_2 = 92\) is the atomic number of uranium, \(e = 1.602 \times 10^{-19} \mathrm{~C}\) is the charge of an electron, and \(E = 5 \mathrm{~MeV} = 5 \times 10^6 \times 1.602 \times 10^{-13} \mathrm{~J}\) is the energy.

Step 3: Substitute the Values into the Formula

Substitute the values into the formula: \(d = \frac{(8.988 \times 10^9) \cdot (2) \cdot (92) \cdot (1.602 \times 10^{-19})^2}{2 \cdot (5 \times 10^6 \times 1.602 \times 10^{-13})}\).

Step 4: Calculate the Distance of Closest Approach

Perform the calculations step-by-step:1. Calculate \(e^2 = (1.602 \times 10^{-19})^2 = 2.566 imes 10^{-38} \mathrm{~C^2}\).2. Calculate the numerator: \((8.988 \times 10^9) \times 2 \times 92 \times 2.566 \times 10^{-38} = 4.235 imes 10^{-26} \mathrm{~J~m}\).3. Calculate the denominator: \(2 \times (5 \times 10^6 \times 1.602 \times 10^{-13}) = 1.602 \times 10^{-6} \mathrm{~J}\).4. Compute \(d = \frac{4.235 \times 10^{-26}}{1.602 \times 10^{-6}} = 2.644 \times 10^{-20} \mathrm{~m}\).

Step 5: Convert the Distance to Convenient Units

Convert the distance from meters to centimeters: \(2.644 \times 10^{-20} \mathrm{~m} = 2.644 \times 10^{-18} \mathrm{~cm}\). This value is closest to option \(d\), which is \(10^{-15} \mathrm{~cm}\), so the answer most aligns with statement \(d\).

Key Concepts

Coulomb's Law

Coulomb's Law is the foundation for understanding the interactions between charged particles. It expresses the force between two charges as directly proportional to the product of their charges and inversely proportional to the square of the distance between them.This force can be attractive or repulsive, depending on the nature of the charges. The formula for this interaction is given by:\[F = k_e \cdot \frac{{|q_1 \cdot q_2|}}{{r^2}}\]where

• \(F\) is the magnitude of the force between the charges,
• \(k_e\) is Coulomb's constant \((8.988 \times 10^9 \text{ N m}^2/ ext{C}^2)\),
• \(q_1\) and \(q_2\) are the magnitudes of the charges,
• and \(r\) is the distance between the centers of the two charges.

This concept is crucial in scenarios like the problem at hand, where the \(\alpha\)-particle approaches a fixed uranium nucleus. The force and energy interactions are governed by this fundamental principle.

Alpha Particle Scattering

Alpha particle scattering is a significant phenomenon in nuclear physics experiments, particularly famous because of Rutherford's gold foil experiment. This type of scattering involves high-energy alpha particles (consisting of two protons and two neutrons) directed at a thin sheet of metal foil. The way these particles are deflected can provide insight into the structure of the atom.In the exercise mentioned, when an alpha particle is scattered by a uranium nucleus, it experiences deflection due to the electrostatic forces described by Coulomb's law.

• This can lead to a 'distance of closest approach', where the particle's forward momentum is fully countered by Coulombic repulsion.
• The scattering angle, such as the given \(180^\circ\), signifies a complete reversal in the motion of the alpha particle, indicating how close the particle gets to the nucleus before being repelled.

Understanding alpha particle scattering helps illustrate the nucleus's size and charge distribution, showing that most of an atom's mass is concentrated in a small central nucleus.

Nuclear Physics

Nuclear Physics is the branch of physics dealing with the components and structure of the atomic nucleus. This field is fundamental to understanding nuclear reactions, forces, and particles like alpha particles, composing the nucleus. When dealing with problems like analyzing the scattering of an alpha particle, nuclear physics explores:

• The behavior of subatomic particles under various force fields,
• The energy transformations occurring during interactions, and
• The insights these interactions provide about the structure and characteristics of the nuclei involved.

The phenomenal ability of Rutherford to derive the nuclear model of the atom came from analyzing data from alpha particle scattering experiments — an application of nuclear physics that illustrated the small but dense nature of an atom's nucleus.

Electrostatic Potential Energy

Electrostatic potential energy arises in charged particle systems due to the forces exerted by electric fields. For any two-point charges, this potential energy is part of the energy conservation principle, where kinetic energy can be transformed into potential energy.In our exercise, when an alpha particle approaches a uranium nucleus, its kinetic energy gradually converts to electrostatic potential energy.This energy transformation allows us to compute the 'distance of closest approach'. This is when the particle stops moving closer, as all its kinetic energy is converted.The potential energy \(U\) between two charges is given by:\[U = k_e \cdot \frac{{q_1 \cdot q_2}}{{r}}\]where

• \(k_e\) is Coulomb's constant,
• \(q_1\) and \(q_2\) are the magnitudes of the charges,
• and \(r\) is the separation distance.

This equation reflects how potential energy is inversely proportional to distance — meaning, as the particle gets closer, the potential energy increases, and kinetic diminishes, a critical point in solving distance of closest approach problems.