/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 80 A proton and an alpha particle a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A proton and an alpha particle are released from rest when they are 0.225 nm apart. The alpha particle (a helium nucleus) has essentially four times the mass and two times the charge of a proton. Find the maximum speed and maximum acceleration of each of these particles. When do these maxima occur, just following the release of the particles or after a very long time?

Short Answer

Expert verified
Max acceleration occurs at release; max speed after long time.

Step by step solution

01

Understanding the Problem

We have a proton and an alpha particle initially at rest, positioned 0.225 nm apart. The alpha particle has four times the mass and twice the charge of the proton. To find their maximum speed and acceleration, we need to consider the electric forces acting on them as they move apart due to their mutual electrostatic repulsion.
02

Calculating Electric Force

The electric force can be calculated using Coulomb's Law: \[ F = k \frac{|q_1 q_2|}{r^2} \]where \( q_1 = e \) (charge of proton), \( q_2 = 2e \) (charge of alpha particle), and \( r = 0.225 \times 10^{-9} \text{ m} \). The constant \( k \approx 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \). Substitute these values to find the initial force between the particles.
03

Calculating Maximum Acceleration

The maximum acceleration occurs when the particles just begin to move, as forces are largest at the closest distance. Use Newton's second law:\[ a = \frac{F}{m} \]for both the proton and the alpha particle. Calculate their accelerations using masses: \( m_p = 1.67 \times 10^{-27} \text{ kg} \) for the proton and \( m_{\alpha} = 4 \times m_p \) for the alpha particle.
04

Maximum Speed and Energy Conservation

Using conservation of energy, initial potential energy due to electrostatic force equals the kinetic energy as they move far apart:\[ k \frac{|q_1 q_2|}{r} = \frac{1}{2} m_p v_p^2 + \frac{1}{2} m_{\alpha} v_{\alpha}^2 \]Substitute known values to solve for maximum speeds \( v_p \) and \( v_{\alpha} \). Consider the charges and masses for each particle in these calculations.
05

When do the Maxima Occur?

The maximum acceleration occurs immediately as they begin to move. The maximum speeds occur when they are far apart, as initially stored potential energy is fully converted into kinetic energy when the particles are sufficiently distanced so forces are negligible.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Coulomb's Law
Coulomb's Law is a fundamental principle of electrostatics that describes how the electric force between two charged objects behaves. This law states that the force (\[ F \]) between two point charges is directly proportional to the product of the magnitude of the charges (\[ |q_1 q_2| \]) and inversely proportional to the square of the distance (\[ r^2 \]) between them. The equation is given by:
  • \( F = k \frac{|q_1 q_2|}{r^2} \)
  • Where \( k \) is the electrostatic constant, approximately \( 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \).
This principle explains the repulsive force experienced by both the proton and the alpha particle in the exercise, causing them to move apart. As their charges are positive, they repel each other, and the magnitude of this repulsion diminishes as they move further away.
Understanding Coulomb's Law is key to predicting how charged particles will interact, influencing both the force they experience and their subsequent motion.
Newton's Second Law
Newton's Second Law of Motion provides insight into how forces affect the motion of objects. According to this law, the acceleration (\[ a \]) of an object is directly proportional to the net force (\[ F \]) acting upon it and inversely proportional to its mass (\[ m \]). This relationship is expressed by the equation:
  • \( a = \frac{F}{m} \)
In the given exercise, Newton's Second Law helps us calculate the maximum acceleration of both the proton and the alpha particle. Initially, both particles experience the greatest force at the closest distance when they are about to start moving.
By knowing the masses of each particle (with the proton's known mass and the alpha particle being four times heavier), we can use this formula to find how quickly each particle starts to accelerate.
This concept is essential for understanding how different masses will react differently when subjected to the same force, as seen with our proton and alpha particle.
Energy Conservation
The principle of energy conservation is a cornerstone of physics, stating that energy cannot be created or destroyed in an isolated system. Instead, it transforms from one form to another. In our scenario, we consider the transformation between potential and kinetic energy.
  • Initially, the potential energy of the system is given by the electrostatic potential energy:\( k \frac{|q_1 q_2|}{r} \).
  • As the charges move apart, this potential energy is converted to kinetic energy:\( \frac{1}{2} m_p v_p^2 + \frac{1}{2} m_{\alpha} v_{\alpha}^2 \).
The maximum speeds of the proton and alpha particle are achieved when the electrostatic potential energy is fully converted into kinetic energy.
This marks the condition where they are far enough apart for the forces between them to become negligible.
Understanding how energy conservation influences the motion of charged particles is crucial in various scientific and engineering applications.
Acceleration
Acceleration is the change in velocity of an object over time, commonly described in meters per second squared (\[ m/s^2 \]). It plays a critical role in determining how quickly an object speeds up or slows down.
  • The initial maximum acceleration occurs at the point of release when the force is at its peak.
  • By applying Newton's Second Law, \( a = \frac{F}{m} \), we ascertain the particles' accelerations.
In this exercise, the forces acting on the particles are strongest when they are closest. Thus, their acceleration is greatest right after they are released.
As they move apart, both the force and resulting acceleration steadily decrease.
This concept helps us appreciate how forces and mass interact to dictate the motion of objects, pivotal in both theoretical and practical physics applications.
Kinetic Energy
Kinetic Energy is the energy that an object possesses due to its motion and is expressed mathematically as:
  • \( KE = \frac{1}{2} m v^2 \)
This form of energy reflects how much work an object can do due to its movement.
In the context of this exercise, as the proton and the alpha particle move away from their initial positions, the potential energy is transformed into kinetic energy.
This transition highlights the maximum kinetic energy state when the particles are sufficiently distanced such that the remaining forces are negligible.
Understanding kinetic energy, especially in systems involving charged particles, is fundamental for evaluating movement and collisions in physics and engineering domains.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A charge of 28.0 \(\mathrm{nC}\) is placed in a uniform electric field that is directed vertically upward and that has a magnitude of \(4.00 \times 10^{4} \mathrm{N} / \mathrm{C}\) . What work is done by the electric force when the charge moves (a) 0.450 \(\mathrm{m}\) to the right; (b) 0.670 \(\mathrm{m}\) upward; (c) 2.60 \(\mathrm{m}\) at an angle of \(45.0^{\circ}\) downward from the horizontal?

Electric eels. Electric eels and electric fish generate large potential differences that are used to stun 9 enemies and prey. These potentials are produced by cells that each can generate 0.10 V. We can plausibly model such cells as charged capacitors. (a) How should these cells be connected \((\mathrm{in}\) series or in parallel) to produce a total potential of more than 0.10 \(\mathrm{V} ?\) (b) Using the connection in part (a), how many cells must be connected together to produce the 500 \(\mathrm{V}\) surge of the electric eel?

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.)

A \(\mathrm{A} 20.0 \mu \mathrm{F}\) capacitor is charged to a potential difference of 800 \(\mathrm{V} .\) The terminals of the charged capacitor are then connected to those of an uncharged 10.0\(\mu \mathrm{F}\) capacitor. Compute (a) the onginal charge of the system, (b) the final potential difference across each capacitor, (c) the final energy of the system, and (d) the decrease in energy when the capacitors are connected.

Potential in human cells. Some cell walls in the human body have a layer of negative charge on the inside surface and a layer of positive charge of equal magnitude on the outside surface. Suppose that the charge density on either surface is \(\pm 0.50 \times 10^{-3} \mathrm{C} / \mathrm{m}^{2},\) the cell wall is 5.0 \(\mathrm{nm}\) thick, and the cell-wall material is air. (a) Find the magnitude of \(\vec{E}\) in the wall between the two layers of charge. (b) Find the potential difference between the inside and the outside of the cell. Which is at the higher potential? (c) A typical cell in the human body has a volume of \(10^{-16} \mathrm{m}^{3} .\) Estimate the total electric-field energy stored in the wall of a cell of this size. (Hint: Assume that the cell is spherical, and calculate the volume of the cell wall.) (d) In reality, the cell wall is made up, not of air, but of tissue with a dielectric constant of \(5.4 .\) Repeat parts (a) and (b) in this case.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.