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Chapter 23: Problem 74

An alpha particle with kinetic energy 9.50 MeV (when far away) collides head- on with a lead nucleus at rest. What is the distance of closest approach of the two particles? (Assume that the lead nucleus remains stationary and may be treated as a point charge. The atomic number of lead is 82. The alpha particle is a helium nucleus, with atomic number 2.)

Short Answer

Expert verified
The distance of closest approach is approximately \( 3.00 \times 10^{-14} \text{ m} \).

Step by step solution

01

Identify the Charges and Initial Conditions

An alpha particle, which is a helium nucleus, has an atomic number of 2, meaning its charge is \( q_1 = 2e \). The lead nucleus has an atomic number of 82, thus its charge is \( q_2 = 82e \). The kinetic energy of the alpha particle is given as 9.50 MeV when far apart from the lead nucleus.
02

Use Conservation of Energy

Initially, the system's energy comprises only the kinetic energy of the alpha particle, since the potential energy is negligible when they are far apart. As the alpha particle approaches the lead nucleus, its kinetic energy is converted to electric potential energy. At the closest approach, all the kinetic energy has converted into potential energy: \[ \frac{1}{4\pi\varepsilon_0} \cdot \frac{q_1 q_2}{r} = K \] where \( K = 9.50 \text{ MeV} \) and \( r \) is the distance of closest approach.
03

Convert Units and Constants

Convert the kinetic energy from MeV to joules. \[ 1 \text{ MeV} = 1.602 \times 10^{-13} \text{ J} \] So, \( K = 9.50 \times 1.602 \times 10^{-13} \text{ J} \). The charge of an electron, \( e = 1.602 \times 10^{-19} \text{ C} \), and the Coulomb's constant, \( \frac{1}{4\pi\varepsilon_0} = 8.988 \times 10^9 \text{ N m}^2/\text{C}^2 \).
04

Solve for the Closest Approach Distance

Substitute known values into the energy equation:\[ \frac{(8.988 \times 10^9) \cdot (2 \times 1.602 \times 10^{-19}) \cdot (82 \times 1.602 \times 10^{-19})}{r} = 9.50 \times 1.602 \times 10^{-13} \] Solve for \( r \):\[ r = \frac{8.988 \times 10^9 \times (2 \times 1.602 \times 10^{-19}) \times (82 \times 1.602 \times 10^{-19})}{9.50 \times 1.602 \times 10^{-13}} \]
05

Calculate the Numeric Solution

Now compute the expression:\[ r \approx \frac{8.988 \times 10^9 \times (2 \times 1.602 \times 10^{-19}) \times (82 \times 1.602 \times 10^{-19})}{9.50 \times 1.602 \times 10^{-13}} \approx 3.00 \times 10^{-14} \text{ m} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Alpha Particle Collision
An alpha particle collision is an intriguing event in nuclear physics. Alpha particles are helium nuclei, consisting of two protons and two neutrons, making them relatively massive among particles. This massiveness, combined with their positive charge (due to the two protons), means they can have significant interactions with other nuclei.
When an alpha particle collides head-on with a nucleus like lead, it interacts through the Coulomb force, preventing direct contact. The strong positive charges repel each other, stopping the alpha particle at a certain distance, known as the closest approach. Understanding this collision helps illuminate the behavior of subatomic particles and nuclear forces.
Conservation of Energy
The principle of conservation of energy is crucial in physics and plays a vital role in analyzing alpha particle collisions. This principle states that energy cannot be created or destroyed, only transformed from one form to another.
  • In the context of particle collision, the initial kinetic energy of the alpha particle must end as potential energy at the closest approach point.
  • This energy transformation allows us to find the distance of closest approach, as the particle's initial kinetic energy directly equals the electric potential energy at closest separation.
This concept assures us that despite the interaction, the total energy remains constant.
Electric Potential Energy
Electric potential energy is the energy stored due to the position of charged particles in an electric field. In an alpha particle collision, this potential energy becomes significant as the particles approach each other.
  • Initially, when the particles are far apart, the electric potential energy is negligible.
  • As the alpha particle nears the nucleus, its kinetic energy converts into electric potential energy.
  • The potential energy can be calculated using the formula: \[ U = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q_1 q_2}{r} \]
At the closest approach, all initial kinetic energy is converted into potential energy, allowing us to calculate the distance of closest approach precisely. Understanding this energy shift is vital for studying particle interactions.
Coulomb's Law
Coulomb's law is the fundamental principle describing the electrostatic interaction between electrically charged particles. It quantifies the force of attraction or repulsion between two charged bodies.
  • The law is expressed as: \[ F = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q_1 q_2}{r^2} \]
  • Here, \( F \) is the force between charges, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, \( r \) is the separation distance, and \( \varepsilon_0 \) is the permittivity of free space.
In the context of alpha particle collisions, Coulomb's law describes the repulsive force that temporarily halts the alpha's advancement, allowing for the calculation of the closest approach. It highlights the balance of forces and energy transformations essential in subatomic interactions.

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Most popular questions from this chapter

A very large plastic sheet carries a uniform charge density of \(-\)6.00 nC\(/\)m\(^2\) on one face. (a) As you move away from the sheet along a line perpendicular to it, does the potential increase or decrease? How do you know, without doing any calculations? Does your answer depend on where you choose the reference point for potential? (b) Find the spacing between equipotential surfaces that differ from each other by 1.00 V. What type of surfaces are these?

In a certain region of space the electric potential is given by \(V = +Ax^2y - Bxy^2,\) where \(A =\) 5.00 \(V/m^3\) and \(B =\) 8.00 \(V/m^3\). Calculate the magnitude and direction of the electric field at the point in the region that has coordinates \(x =\) 2.00 m, \(y =\) 0.400 m, and \(z = 0\).

A positive charge \(q\) is fixed at the point \(x = 0, y = 0\), and a negative charge \(-2_q\) is fixed at the point \(x = a, y = 0\). (a) Show the positions of the charges in a diagram. (b) Derive an expression for the potential \(V\) at points on the \(x\)-axis as a function of the coordinate \(x\). Take \(V\) to be zero at an infinite distance from the charges. (c) At which positions on the \(x\)-axis is \(V = 0\)? (d) Graph \(V\) at points on the \(x\)-axis as a function of \(x\) in the range from \(x = -2a\) to \(x = +2a\). (e) What does the answer to part (b) become when \(x \gg a\)? Explain why this result is obtained.

Two stationary point charges \(+\)3.00 nC and \(+\)2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at a point midway between the two charges and moves along the line connecting the two charges. What is the speed of the electron when it is 10.0 cm from the \(+\)3.00-nC charge?

Two plastic spheres, each carrying charge uniformly distributed throughout its interior, are initially placed in contact and then released. One sphere is 60.0 cm in diameter, has mass 50.0 g, and contains \(-\)10.0 \(\mu\)C of charge. The other sphere is 40.0 cm in diameter, has mass 150.0 g, and contains \(-\)30.0 \(\mu\)C of charge. Find the maximum acceleration and the maximum speed achieved by each sphere (relative to the fixed point of their initial location in space), assuming that no other forces are acting on them. (\(Hint:\) The uniformly distributed charges behave as though they were concentrated at the centers of the two spheres.)
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