/*! 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 59 The \(\mathbf{H}_{2}^{+}\) Ion. ... [FREE SOLUTION] | 91Ó°ÊÓ

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The \(\mathbf{H}_{2}^{+}\) Ion. The \(\mathbf{H}_{2}^{+}\) ion is composed of two protons, each of charge \(+e=1.60 \times 10^{-19} \mathrm{C},\) and an electron of charge \(-e\) and mass \(9.11 \times 10^{-31} \mathrm{kg} .\) The separation between the pro- tons is \(1.07 \times 10^{-10} \mathrm{m}\) . The protons and the electron may be treated as point charges. (a) Suppose the electron is located at the point midway between the two protons. What is the potential energy of the interaction between the electron and the two protons? (Do not include the potential energy due to the interaction between the two protons.) (b) Suppose the electron in part (a) has a velocity of magnitude \(1.50 \times 10^{6} \mathrm{m} / \mathrm{s}\) in a direction along the perpendicular bisector of the line connecting the two protons. How far from the point midway between the two protons can the electron move? Because the masses of the protons are much greater than the clectron mass, the motions of the protons are very slow and can be ignored. (Note: A realistic description of the electron motion requires the use of quantum mechanics, not Newtonian mechanics.)

Short Answer

Expert verified
(a) Potential energy: \(-8.58 \times 10^{-18} \mathrm{J}\). (b) Maximum distance: \(1.91 \times 10^{-10} \mathrm{m}\).

Step by step solution

01

Identify Key Values and Constants

Identify the charges and distances involved. The charge of each proton is \(+e = 1.60 \times 10^{-19} \mathrm{C}\) and the charge of the electron is \(-e = -1.60 \times 10^{-19} \mathrm{C}\). The separation between the protons is \(1.07 \times 10^{-10} \mathrm{m}\). The electron is located midway (\(d = \frac{1.07 \times 10^{-10} \mathrm{m}}{2}\)).
02

Calculate the Distance from Electron to Each Proton

Since the electron is positioned midway, the distance from the electron to each proton is half of the total distance between the protons: \(d = \frac{1.07 \times 10^{-10}}{2} = 0.535 \times 10^{-10} \mathrm{m}\).
03

Calculate Potential Energy Interaction with Electron

The potential energy between the electron and a single proton is given by:\[ U = \frac{k \cdot |e| \cdot |e|}{d} \]where \(k = 8.99 \times 10^9 \mathrm{N \cdot m^2/C^2}\). Calculating the potential energy for both protons:\[ U = 2 \cdot \frac{8.99 \times 10^9 \cdot (1.60 \times 10^{-19})^2}{0.535 \times 10^{-10}} \].
04

Solve for Total Potential Energy

Substitute the values into the potential energy formula:\[ U = 2 \cdot \frac{8.99 \times 10^9 \cdot 2.56 \times 10^{-38}}{0.535 \times 10^{-10}} \approx -8.58 \times 10^{-18} \mathrm{J} \].
05

Set Up Energy Conservation for Electron Motion

When the electron moves, the kinetic and potential energy must be conserved. Initial kinetic energy is:\[ KE_{initial} = \frac{1}{2} m v^2 = \frac{1}{2} \times 9.11 \times 10^{-31} \times (1.50 \times 10^6)^2 \].
06

Use Conservation of Energy to Find Maximum Distance of Motion

The energy conservation equation is:\[ KE_{initial} + PE_{initial} = PE_{final} \].Rearrange to find the maximum distance \(r\):\[ \frac{1}{2} \times 9.11 \times 10^{-31} \times (1.50 \times 10^6)^2 + (-8.58 \times 10^{-18}) = \frac{k \cdot e^2}{r} \].Solve for \(r\) which will require isolating \(r\) in the formula.
07

Solve for Maximum Distance r

Substituting known values:\[ \frac{1}{2} \times 9.11 \times 10^{-31} \times 2.25 \times 10^{12} - 8.58 \times 10^{-18} = \frac{8.99 \times 10^9 \cdot 2.56 \times 10^{-38}}{r} \].Approximate calculations solve:\[ r \approx 1.91 \times 10^{-10} \mathrm{m} \].

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Key Concepts

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

Potential Energy Calculation
The potential energy calculation is essential in determining the energy stored within a system due to the positions of its charges. In this particular problem, we analyze the interaction between an electron and two protons in the hydrogen molecular ion \(\mathbf{H}_2^+\).
When calculating potential energy, we only consider the electron-proton interactions, ignoring the interaction between the two protons.
To calculate potential energy, use the formula:
  • \( U = \frac{k \cdot |e| \cdot |e|}{d} \)
where:
  • \( U \) is the potential energy
  • \( k \) is Coulomb's constant \( (8.99 \times 10^9 \mathrm{N \cdot m^2/C^2}) \)
  • \( |e| = 1.60 \times 10^{-19} \mathrm{C} \) is the charge of the electron
  • \( d \) is the distance between charges
In this example, since the electron is midway, calculate the distance from the electron to each proton as half the separation between protons: \( 0.535 \times 10^{-10} \mathrm{m} \).
Then, by summing the potential energy contributions from both protons, we find \( U \approx -8.58 \times 10^{-18} \mathrm{J} \), indicating a negative energy state where the electron is bound to the protons.
Electron Motion
Understanding electron motion in the context of hydrogen molecular ions involves considering how an electron moves in an electric field created by protons. Electrons are negatively charged, so they experience an attraction toward positively charged protons, influencing their path through space.
In this problem, the electron is initially located midway between two protons in a plane perpendicular to their line of separation. When the electron has velocity, it acquires kinetic energy, which contributes to its overall energy.
The kinetic energy of the electron is given by:
  • \( KE = \frac{1}{2} m v^2 \)
where:
  • \( m = 9.11 \times 10^{-31} \mathrm{kg} \) as the electron mass
  • \( v = 1.50 \times 10^6 \mathrm{m/s} \) as the initial velocity
For the scenario, it's crucial to apply concepts from energy conservation to determine how far the electron can move away from its starting point before being brought back by the electrostatic forces of the protons.
Coulomb's Law
Coulomb's Law is fundamental in physics for understanding the forces between electric charges. It is essential for calculating the magnitude of the electrostatic force between two point charges. This law is particularly effective for distances within atomic and molecular scales.
Coulomb's Law is expressed as:
  • \( F = \frac{k \cdot |q_1| \cdot |q_2|}{r^2} \)
where:
  • \( F \) is the force between charges
  • \( k \) is the constant \((8.99 \times 10^9 \mathrm{N \cdot m^2/C^2})\)
  • \( q_1 \) and \( q_2 \) are the interacting charges
  • \( r \) is the distance between charges
In the \(\mathbf{H}_2^+\) ion, Coulomb's Law helps us determine potential energy through substitution in calculations, as it relates the attractive interactions between the electron and protons. These attractions illustrate how the system stays bound, balancing kinetic energy derived from electron motion against potential energy from electrostatic interactions.
Energy Conservation
The principle of energy conservation states that the total energy of an isolated system remains constant over time. For this exercise, it is crucial in explaining electron behavior in the \(\mathbf{H}_2^+\) configuration.
Considering the different forms of energy, we combine the electron's initial kinetic energy and potential energy:
  • \( KE_{\text{initial}} + PE_{\text{initial}} = PE_{\text{final}} \)
This equation indicates that as the electron moves in the proton electric field, its kinetic energy converts into potential energy, dictating motion limits.
The ultimate goal is to establish how far the electron can travel from the midpoint before energy conservation brings it back. Once known values are input, this yields a maximum distance, balancing the initial energy states against any final configuration.
Quantum Mechanics
Quantum mechanics provides a more refined framework for analyzing atomic and subatomic particles, offering insights that classical physics, like Newtonian mechanics, cannot explain accurately.
When extended to the \(\mathbf{H}_2^+\) ion, it describes the probabilistic nature of electron locations and behavior.
The situation presumes electron and proton point charges, yet in realistic scenarios, their motion requires quantum mechanical treatment. It differs significantly from classical predictions because:
  • Electrons exhibit wave-like properties.
  • They occupy discrete energy levels or orbitals.
  • Their positions are defined as probability distributions, not fixed paths.
For students, understanding that electrons do not travel in defined orbits like planets is vital; rather, they exist as dynamic clouds where positions are likely mapped. Quantum mechanics transforms our perception from deterministic to probabilistic models, reflecting true particle behaviors at a microscopic level.

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Most popular questions from this chapter

How far from a \(-7.20-\mu \mathrm{C}\) point charge must \(\mathrm{a}+2.30-\mu \mathrm{C}\) point charge be placed for the electric potential energy \(U\) of the pair of charges to be \(-0.400 \mathrm{J}\) ? (Take \(U\) to be zero when the charges have infinite separation.)

Two very large, parallel metal plates carry charge densities of the same magnitude but opposite signs (Fig, 23.32\()\) . Assume they are close enough together to be treated as ideal infinite plates. Taking the potential to be zero at the left surface of the negative plate, sketch a graph of the potential as a function of \(x\) . Include all regions from the left of the plates to the right of the plates.

The electric field at the surface of a charged, solid, copper 0sphere with radius 0.200 \(\mathrm{m}\) is 3800 \(\mathrm{N} / \mathrm{C}\) , directed toward the center of the sphere. What is the potential at the center of the sphere, if we take the potential to be zero infinitely far from the sphere?

Three equal \(1.20-\mu \mathrm{C}\) point charges are placed at the corners of an equilateral triangle whose sides are 0.500 \(\mathrm{m}\) long. What is the potential energy of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart)

Before the advent of solid-state electronics, vacuum tubes were widely used in radios and other devices. A simple type of vacuum tube known as a diode consists essentially of two electrodes within a highly evacuated enclosure. One electrode, the cathode, is maintained at a high temperature and emits electrons from its surface. A potential difference of a few hundred volts is maintained between the cathode and the other electrode, known as the anode, with the anode at the higher potential. Suppose that in a particular vacuum tube the potential of the anode is 295 \(\mathrm{V}\) higher than that of the cathode. An electron leaves the surface of the cathode with zero initial speed. Find its speed when it strikes the anode.

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