/*! 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 14 An electron and a proton are ini... [FREE SOLUTION] | 91Ó°ÊÓ

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An electron and a proton are initially very far apart (effectively an infinite distance apart). They are then brought together to form a hydrogen atom, in which the electron orbits the proton at an average distance of \(5.29 \times 10^{-11} \mathrm{m} .\) What is EPE final \(-\mathrm{EPE}_{\text { initial }},\) which is the change in the electric potential energy?

Short Answer

Expert verified
The change in electric potential energy is \(-2.18 \times 10^{-18} \text{ J}.\)

Step by step solution

01

Understand the Concept

The electric potential energy (EPE) between two charged particles is determined by Coulomb's law. The formula for EPE is \( EPE = \frac{k \cdot q_1 \cdot q_2}{r} \), where \( k \) is Coulomb's constant \( (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) \), \( q_1 \) and \( q_2 \) are the charges of the electron and proton respectively, and \( r \) is the separation distance between them. Initially, they are at an infinite distance apart, resulting in an initial EPE of 0.
02

Identify the Charges and Distance

The charge of a proton, \( q_p \), is \( 1.6 \times 10^{-19} \text{ C} \) and the charge of an electron, \( q_e \), is \( -1.6 \times 10^{-19} \text{ C} \). The average distance \( r \) when they form a hydrogen atom is \( 5.29 \times 10^{-11} \text{ m} \).
03

Calculate Final Electric Potential Energy

Now we will find the final EPE using the formula: \[ EPE = \frac{k \cdot q_p \cdot q_e}{r} = \frac{(8.99 \times 10^9 \text{ N m}^2/\text{C}^2) \times (1.6 \times 10^{-19} \text{ C}) \times (-1.6 \times 10^{-19} \text{ C})}{5.29 \times 10^{-11} \text{ m}}. \]
04

Simplify the Calculation

Plug in the values into the equation and compute: \[ EPE = \frac{(8.99 \times 10^9) \times (1.6 \times 10^{-19})^2}{5.29 \times 10^{-11}}. \] Evaluating this gives: \[ EPE = -2.18 \times 10^{-18} \text{ J}. \]
05

Find the Change in Electric Potential Energy

Since the initial EPE is zero (particles at infinity), the change in EPE, \( \Delta EPE \), is simply the final EPE value. Therefore, \[ \Delta EPE = EPE_{final} - EPE_{initial} = -2.18 \times 10^{-18} \text{ J} - 0. \] Hence, \( \Delta EPE = -2.18 \times 10^{-18} \text{ J}. \)

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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 that describes the force between two charged particles. This force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. The formula is given by:
\[ F = \frac{k \cdot q_1 \cdot q_2}{r^2} \]where:
  • \( F \) is the force between the charges
  • \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \)
  • \( q_1 \) and \( q_2 \) are the magnitudes of the charges
  • \( r \) is the separation distance between the charges.
When dealing with electric potential energy (EPE), we modify this formula slightly to account for potential energy, leading to:\[ EPE = \frac{k \cdot q_1 \cdot q_2}{r}. \] Understanding this law is key to analyzing the behavior of charged particles, such as electrons and protons in a hydrogen atom formation.
hydrogen atom formation
The formation of a hydrogen atom is a fundamental process that involves an electron and a proton coming together. Initially, these particles start infinitely far apart, where their interaction is negligible. As they approach each other, they begin to experience significant electric forces, eventually forming a hydrogen atom.
In a hydrogen atom, the electron orbits the proton at an average distance, known as the Bohr radius, which is approximately \(5.29 \times 10^{-11} \text{ m}\). During this process, the potential energy of the system changes from zero (when they are infinitely apart) to a negative value, indicating a stable, bound state.
This change in electric potential energy signifies that the atom is in a lower energy state, having released energy to the surroundings or converting it into kinetic energy of the electron as it settles into orbit around the proton.
charge of electron and proton
The electron and proton are fundamental subatomic particles. They carry charges that have the same magnitude but opposite in sign.
  • Electron: The charge is negative and is valued at \(-1.6 \times 10^{-19} \text{ C} \).
  • Proton: The charge is positive and is valued at \(+1.6 \times 10^{-19} \text{ C} \).
These opposite charges are what allow the electron to orbit the proton in a hydrogen atom, as they exert attractive forces on each other. This attraction is essential for the stability of the hydrogen atom and overall atomic structure.
The interplay of these charges underlies the relationship described by Coulomb's law and plays a vital role in determining the electric potential energy when the electron and proton are at a certain separation distance.
separation distance
The separation distance, often denoted as \( r \), between charged particles is crucial in determining the strength of the electric force and potential energy between them. In the context of a hydrogen atom, this is the average distance between the electron and the proton, called the Bohr radius.
For the hydrogen atom, this distance is \(5.29 \times 10^{-11} \text{ m}\). At this distance:
  • The force and potential energy are just sufficient to keep the electron in orbit around the proton.
  • This configuration represents a stable and minimally energetic state for the hydrogen atom.
The larger the separation distance, the weaker the force and the potential energy, while a smaller separation would increase both. Thus, understanding this balance helps in grasping how atoms maintain their structure and stability.

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Most popular questions from this chapter

Multiple-Concept Example 4 to see the concepts that are pertinent here. In a television picture tube, electrons strike the screen after being accelerated from rest through a potential difference of 25000 \(\mathrm{V}\) . The speeds of the electrons are quite large, and for accurate calculations of the speeds, the effects of special relativity must be taken into account. Ignoring such effects, find the electron speed just before the electron strikes the screen.

Equipotential surface \(A\) has a potential of \(5650 \mathrm{V},\) while equipotential surface \(B\) has a potential of 7850 \(\mathrm{V}\) . A particle has a mass of \(5.00 \times 10^{-2} \mathrm{kg}\) and a charge of \(+4.00 \times 10^{-5} \mathrm{C}\) . The particle has a speed of 2.00 \(\mathrm{m} / \mathrm{s}\) on surface \(A\) . A nonconservative outside force is applied to the particle, and it moves to surface \(B\) , arriving there with a speed of 3.00 \(\mathrm{m} / \mathrm{s}\) . How much work is done by the outside force in moving the particle from \(A\) to \(B ?\)

A capacitor is constructed of two concentric conducting cylindrical shells. The radius of the inner cylindrical shell is \(2.35 \times 10^{-3} \mathrm{m},\) and that of the outer shell is 2.50 \(\times 10^{-3} \mathrm{m}\) . When the cylinders carry equal and opposite charges of magnitude 1.7 \(\times 10^{-10} \mathrm{C}\) , the electric field between the plates has an average magnitude of \(4.2 \times 10^{4} \mathrm{V} / \mathrm{m}\) and is directed radially outward from the inner shell to the outer shell. Determine (a) the magnitude of the potential difference between the cylindrical shells and \((\text { b) the capacitance of this capacitor. }\)

At a distance of 1.60 \(\mathrm{m}\) from a point charge of \(+2.00 \mu \mathrm{C},\) there is an equipotential surface. At greater distances there additional equipotential surfaces. The potential difference between any two successive surfaces is \(1.00 \times 10^{3} \mathrm{V}\) . Starting at a distance of 1.60 \(\mathrm{m}\) and moving radially outward, how many of the additional equipotential surfaces are crossed by the time the electric field has shrunk to one-half of its initial value? Do not include the starting surface.

During a lightning flash, there exists a potential difference of \(V_{\text { clond }}-V_{\text { ground }}=1.2 \times 10^{9} \mathrm{V}\) between a cloud and the ground. As a result, a charge of \(-25 \mathrm{C}\) is transferred from the ground to the cloud. (a) How much work \(W_{\text { ground-cloud }}\) is done on the charge by the electric force? \((b)\) If the work done by the electric force were used to acceler- ate a \(1100-\mathrm{kg}\) automobile from rest, what would be its final speed? (c) If the work done by the electric force were converted into heat how many kilograms of water at \(0^{\circ} \mathrm{C}\) could be heated to \(100^{\circ} \mathrm{C} ?\)

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