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Chapter 38: Problem 21

A beam of alpha particles is incident on a target of lead. A particular alpha particle comes in "head-on" to a particular lead nucleus and stops \(6.50 \times 10^{-14} \mathrm{m}\) away from the center of the nucleus. (This point is well outside the nucleus.) Assume that the lead nucleus, which has 82 protons, remains at rest. The mass of the alpha particle is \(6.64 \times 10^{-27} \mathrm{kg} .\) (a) Calculate the electrostatic potential energy at the instant that the alpha particle stops. Express your result in joules and in MeV. (b) What initial kinetic energy (in joules and in MeV) did the alpha particle have? (c) What was the initial speed of the alpha particle?

Short Answer

Expert verified
(a) Electrostatic potential energy: \(9.00 \times 10^{-13} \text{J}\) (\(5.62 \text{MeV}\)); (b) Initial kinetic energy: \(9.00 \times 10^{-13} \text{J}\) (\(5.62 \text{MeV}\)); (c) Initial speed: \(1.53 \times 10^7 \text{m/s}\).

Step by step solution

01

Understanding the System

The problem involves calculating the electrostatic potential energy when an alpha particle approaches a lead nucleus and stops. It is crucial to understand that at the stopping point, all the kinetic energy is converted into electrostatic potential energy.
02

Calculating Electrostatic Potential Energy

The electrostatic potential energy (U) between the alpha particle and the lead nucleus can be calculated using the formula:\[ U = \frac{k \cdot q_1 \cdot q_2}{r} \]where \( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \) is the Coulomb's constant, \( q_1 = 2 \cdot e \) is the charge of the alpha particle (\( e = 1.6 \times 10^{-19} \text{C} \)), \( q_2 = 82 \cdot e \) is the charge of the lead nucleus, and \( r = 6.50 \times 10^{-14} \, \text{m} \) is the distance from the center.
03

Substitute Values and Solve for U

Substitute the values into the formula:\[ U = \frac{8.99 \times 10^9 \times (2 \times 1.6 \times 10^{-19}) \times (82 \times 1.6 \times 10^{-19})}{6.50 \times 10^{-14}} \]After calculating, \( U \approx 9.00 \times 10^{-13} \, \text{J} \).
04

Convert U to MeV

To convert the energy from joules to MeV, use the conversion factor \( 1 \, \text{MeV} = 1.602 \times 10^{-13} \, \text{J} \). Thus:\[ U_{\text{MeV}} = \frac{9.00 \times 10^{-13} \, \text{J}}{1.602 \times 10^{-13} \, \text{J/MeV}} \approx 5.62 \, \text{MeV} \]
05

Determine Initial Kinetic Energy

Since the alpha particle stops, its initial kinetic energy must equal the electrostatic potential energy at the stopping point, which is approximately \( 9.00 \times 10^{-13} \, \text{J} \) or \( 5.62 \, \text{MeV} \).
06

Calculate Initial Speed

Use the kinetic energy formula \( KE = \frac{1}{2} m v^2 \) to find the speed:\[ 9.00 \times 10^{-13} = \frac{1}{2} \times 6.64 \times 10^{-27} \times v^2 \]Solving for \( v \), we find \( v \approx 1.53 \times 10^7 \, \text{m/s} \).

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Key Concepts

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

Alpha Particles
Alpha particles are fascinating entities in the realm of nuclear physics. They are essentially helium nuclei, comprising two protons and two neutrons. These particles are positively charged due to the presence of the two protons. This charge plays a significant role in their interactions with other charged bodies. When alpha particles are emitted from radioactive substances, they possess considerable kinetic energy, enabling them to penetrate matter, although not as deeply as beta particles or gamma rays due to their relatively larger mass and charge.
  • Structure: Made of 2 protons and 2 neutrons.
  • Charge: Positively charged.
  • Origin: Emitted from radioactive decay.
In the context of the problem, the alpha particle's kinetic energy is entirely converted into electrostatic potential energy when it approaches the lead nucleus and momentarily stops. This interaction highlights the substantial influence of charge in electrostatic interactions.
Lead Nucleus
The lead nucleus presents an intriguing study due to its complexity and charge. Comprising a significant number of protons and neutrons, the lead nucleus boasts a high atomic number, specifically 82 protons in its structure. This large number of protons leads to a significant positive charge, making it an effective site for interactions with other charged particles like alpha particles.
  • Composition: Holds 82 protons.
  • Charge: Exhibits a significant positive charge.
  • Role: Acts as a stationary target in the exercise.
In the exercise setup, the lead nucleus acts as a stationary target for the alpha particles. As the alpha particle approaches, the positive charge of the lead nucleus interacts electromagnetically, causing a conversion of the alpha particle's kinetic energy into electrostatic potential energy. Understanding the nature of the nucleus helps in comprehending the dynamics of such nuclear interactions.
Coulomb's Law
Coulomb's Law is pivotal in explaining the interactions between charged particles. This law posits that the force between two charged bodies is directly proportional to the product of their charges and inversely proportional to the square of the distance separating them. In mathematical terms, it's expressed as:\[ F = \frac{k \cdot q_1 \cdot q_2}{r^2} \]Where:- \( F \) is the force between the charges,- \( q_1 \) and \( q_2 \) are the charges of the particles,- \( r \) is the distance between the centers of the two charges, and- \( k \) is Coulomb's constant.
  • Proportionalities: Direct with charge product, inverse with distance squared.
  • Application: Used to calculate electrostatic potential energy.
In the given problem, Coulomb's Law forms the backbone for calculating the electrostatic potential energy between the alpha particle and the lead nucleus. This formula transforms the problem of force into one of energy by integrating distance and charge into a convenient scalar form of energy.
Kinetic Energy Conversion
Kinetic energy conversion is a crucial concept in understanding energy transformations. In the context of the given exercise, the conversion between kinetic energy and electrostatic potential energy takes center stage. When the alpha particle moves towards the lead nucleus, it possesses kinetic energy owing to its motion. Upon approaching the lead nucleus, this kinetic energy is gradually converted into potential energy due to the electrostatic force acting between the alpha particle and the nucleus.
  • Initial Possession: Alpha particle has kinetic energy when in motion.
  • Conversion: Kinetic energy is transformed into potential energy.
  • Equivalence: The initial kinetic energy equals the electrostatic potential energy at the stopping point.
The exercise vividly demonstrates the profound principle that energy, while changing forms, is always conserved within an isolated system. By calculating the stopping distance, the concrete measure of potential energy is determined, showing a balance between initial kinetic and resultant potential energies.

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Most popular questions from this chapter

Protons are accelerated from rest by a potential difference of 4.00 \(\mathrm{kV}\) and strike a metal target. If a proton produces one photon on impact, what is the minimum wavelength of the resulting x rays? How does your answer compare to the minimum wave-length if \(4.00-\mathrm{keV}\) electrons are used instead? Why do x-ray tubes use electrons rather than protons to produce x rays? use electrons rather than protons to produce x rays?

A photon scatters in the backward direction \(\left(\theta=180^{\circ}\right)\) from a free proton that is initially at rest. What must the wave- length of the incident photon be if it is to undergo a 10.0\(\%\) change in wavelength as a result of the scattering?

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