Problem 78 - College Physics

Question

An alpha particle with a kinetic energy of 10.0 MeV makes a head-on collision with a gold nucleus at rest. What is the distance of closest approach of the two particles? (Assume that the gold nucleus remains stationary and that it may be treated as a point charge. The atomic number of gold is \(79,\) and an alpha particle is a helium nucleus consisting of two protons and two neutrons.)

Step-by-Step Solution

Answer

The distance of closest approach is approximately 34.7 femtometers.

Step 1: Understand the Problem

The problem asks us to find the distance of closest approach between an alpha particle and a gold nucleus. This is a point in space where the kinetic energy of the alpha particle is completely converted into electrostatic potential energy due to the repulsion between the positively charged particles.

Step 2: Calculate Charge of Particles

Calculate the charge of the alpha particle and the gold nucleus. The charge of an alpha particle, which is a helium nucleus, is twice the charge of a proton, so \(Q_{\alpha} = 2 \times 1.602 \times 10^{-19} \text{ C}\). The charge of the gold nucleus is the charge of 79 protons, so \(Q_{\text{gold}} = 79 \times 1.602 \times 10^{-19} \text{ C}\).

Step 3: Set Up Energy Conservation Equation

At the distance of closest approach, all the kinetic energy of the alpha particle has been converted into electric potential energy. The initial kinetic energy \(K\) of the alpha particle is \(10.0 \text{ MeV} = 10.0 \times 1.602 \times 10^{-13} \text{ J}\). The potential energy \(U\) at the closest distance \( r \) is given by\[ U = \frac{k Q_{\alpha} Q_{\text{gold}}}{r} \]where \(k = 8.9875 \times 10^9 \text{ N m}^2/\text{C}^2\) is the Coulomb constant.

Step 4: Solve for Closest Approach Distance

Set the initial kinetic energy equal to the potential energy to find \(r\):\[ \frac{k Q_{\alpha} Q_{\text{gold}}}{r} = K \]Rearrange to solve for \(r\):\[ r = \frac{k Q_{\alpha} Q_{\text{gold}}}{K} \]Substitute \(Q_{\alpha}, Q_{\text{gold}}, k, \) and \(K\) into the equation:\[ r = \frac{(8.9875 \times 10^9) (3.204 \times 10^{-19}) (79 \times 1.602 \times 10^{-19})}{10.0 \times 1.602 \times 10^{-13}} \]Calculate \(r\).

Step 5: Final Calculation

Plug the values into the equation and compute:\[ r = \frac{8.9875 \times 10^9 \times 3.204 \times 10^{-19} \times 1.26558 \times 10^{-17}}{1.602 \times 10^{-12}} \]Get the value of \( r \), which is approximately \(34.7 \times 10^{-15} \text{ m} \) or 34.7 femtometers.

Key Concepts

Alpha particle

An alpha particle is a type of ionizing radiation ejected by the nuclei of some unstable atoms. It consists of two protons and two neutrons, making it essentially a helium nucleus. These particles have a positive charge due to the presence of two protons.
This positive charge is significant because it causes an alpha particle to interact with other charged particles through electrostatic forces.

Alpha particles are quite massive compared to other forms of radiation like beta particles and gamma rays. Even though they have low penetration power and can be stopped by just a sheet of paper or a layer of human skin, they're highly effective at ionizing atoms and molecules within that range. Because of this, alpha particles can be hazardous if they are inhaled or ingested.
In the context of physics problems like the one here, their positive charge plays an essential role in understanding how they interact with other charged bodies, like a gold nucleus.

Gold Nucleus

The gold nucleus is composed of 79 protons and typically 118 neutrons, making it a stable, positively charged entity. The atomic number of gold, which is 79, indicates the number of protons within the nucleus, and this directly relates to its charge.
This charge is crucial in calculating interactions with other charged particles, like the alpha particle in this exercise.

• Gold's atomic charge becomes a key factor when determining the electrostatic interaction between the nucleus and another charged particle, since it will determine the strength of the repulsion or attraction experienced.
• Being a highly dense metal, gold atoms have closely packed nuclei, making them relatively stationary during interactions with faster, lighter particles like alpha particles.

In calculations related to atomic interactions, the gold nucleus can often be treated as a point charge due to its significantly larger mass compared to an alpha particle, simplifying the calculation of forces.

Coulomb's law

Coulomb’s law describes the electrostatic interaction between two charged particles. According to this law, the electric force between two charges is proportional to the product of the magnitude of each of the charges and inversely proportional to the square of the distance between them. Mathematically, this is expressed as:
\[ F = k \frac{Q_1 Q_2}{r^2} \]
Here, \( F \) is the force of attraction or repulsion, \( Q_1 \) and \( Q_2 \) are the two charges, \( r \) is the distance between the charges, and \( k \) is Coulomb's constant \( 8.9875 \times 10^9 \, N \cdot m^2/C^2 \).

• Coulomb’s law is vital for calculating the electric potential energy exchanged between charged particles like alpha particles and a gold nucleus.
• It helps us understand the dynamics of energy transformation when calculating the distance of closest approach in problems like these.

The law explains why the alpha particle slows down as it approaches the gold nucleus, as seen in these calculations, because the repulsive electrostatic force between the two increases.

Kinetic energy

Kinetic energy represents the energy of motion that a particle possesses while it is moving. The kinetic energy \( K \) of a particle can be calculated using the equation:
\[ K = \frac{1}{2}mv^2 \]
Usually, for high-speed particles like alpha particles, kinetic energy is often given in units of MeV (million electronvolts) where \( 1 \, MeV = 1.602 \times 10^{-13} \) Joules.

In the context of an alpha particle making a head-on collision with a gold nucleus:

• The kinetic energy initially provides the alpha particle its speed, allowing it to "head-on" approach the gold nucleus.
• As the alpha particle moves increasingly closer to the gold nucleus, its kinetic energy continuously converts into electrostatic potential energy due to the electrostatic repulsion felt more strongly at shorter distances.

This conversion entirely takes place by the distance of closest approach, where the kinetic energy of the alpha particle is zero and all energy has become potential.