Problem 83
Question
The mass of an electron is \(\mathrm{m}\). Its charge is e and it is accelerated from rest through a potential difference \(\mathrm{V}\). The velocity acquired by the electron will be (a) \(\sqrt{\mathrm{V} / \mathrm{m}}\) (b) \(\sqrt{\mathrm{eV} / \mathrm{m}}\) (c) \(\sqrt{2 \mathrm{e} \mathrm{V} / \mathrm{m}}\) (d) none
Step-by-Step Solution
Verified Answer
The velocity acquired by the electron is \( \sqrt{2eV/m} \), option (c).
1Step 1: Understand the Problem
We need to find the velocity acquired by an electron when it is accelerated from rest through a potential difference V. Given parameters are the electron's mass \( m \) and charge \( e \).
2Step 2: Recall the Fundamental Relation
When an electron is accelerated through a potential difference \( V \), its kinetic energy gained is equal to the work done by the electric field. This is given by the equation \( K.E. = eV \), where \( K.E. \) is the kinetic energy acquired by the electron, \( e \) is its charge, and \( V \) is the potential difference.
3Step 3: Express Kinetic Energy in Terms of Velocity
The kinetic energy \( K.E. \) of an electron can also be expressed as \( \frac{1}{2}mv^2 \), where \( m \) is the mass of the electron and \( v \) is its velocity. So, we equate this to the expression for work done: \( \frac{1}{2}mv^2 = eV \).
4Step 4: Solve for Velocity
Rearrange the equation \( \frac{1}{2}mv^2 = eV \) to solve for \( v \):1. Multiply both sides by 2: \( mv^2 = 2eV \).2. Divide both sides by \( m \): \( v^2 = \frac{2eV}{m} \).3. Take the square root of both sides to solve for \( v \): \( v = \sqrt{\frac{2eV}{m}} \).
5Step 5: Select the Correct Option
Compare the expression we derived for velocity, \( v = \sqrt{\frac{2eV}{m}} \), with the given options. Option (c) \( \sqrt{2eV/m} \) matches our expression.
Key Concepts
Kinetic Energy of ElectronPotential Difference AccelerationWork-Energy Principle
Kinetic Energy of Electron
When an electron moves, it holds energy of motion called kinetic energy. This concept is fundamental in understanding how electrons behave when influenced by external forces. For electrons, the kinetic energy is computed using the formula:
If we know the kinetic energy or the velocity, we can determine other associated properties, revealing critical insights into electron motion in various fields, including quantum mechanics and electron microscopy.
Understanding kinetic energy helps us appreciate how electrons gain speed and interact with materials, influencing conduction and other electrical behaviors.
- \[K.E. = \frac{1}{2}mv^2\]Where \( m \) stands for the electron's mass and \( v \) its velocity.
If we know the kinetic energy or the velocity, we can determine other associated properties, revealing critical insights into electron motion in various fields, including quantum mechanics and electron microscopy.
Understanding kinetic energy helps us appreciate how electrons gain speed and interact with materials, influencing conduction and other electrical behaviors.
Potential Difference Acceleration
The acceleration of an electron through a potential difference is a core idea in electromagnetism.
This happens when an electron moves under the influence of an electric field created by a voltage difference. The work done on the electron by this field comes from converting potential energy into kinetic energy.
The relationship between potential difference (\( V \)) and the energy of an electron can be expressed as:
By understanding this concept, we can calculate changes in velocity and control electron motion for different applications, from imaging technologies to particle accelerators.
This happens when an electron moves under the influence of an electric field created by a voltage difference. The work done on the electron by this field comes from converting potential energy into kinetic energy.
The relationship between potential difference (\( V \)) and the energy of an electron can be expressed as:
- \[eV = K.E.\]Where \( e \) is the electron's charge, and \( K.E. \) is the kinetic energy gained.
By understanding this concept, we can calculate changes in velocity and control electron motion for different applications, from imaging technologies to particle accelerators.
Work-Energy Principle
The work-energy principle gives us a powerful way to connect the work done on an electron with changes in its kinetic energy. According to this principle, any work performed on a particle results in an equivalent change in its kinetic energy.
In mathematical terms, when an electron is subjected to forces resulting in a displacement, the work done is expressed as:
It enables the problem-solving mindset needed for developing technologies like renewable energy sources or understanding natural phenomena like lightning. The work-energy principle is more than a mere equation; it's a tool for unlocking how energy influences motion across many contexts.
In mathematical terms, when an electron is subjected to forces resulting in a displacement, the work done is expressed as:
- #The work is given by \( W = eV \), which implies the energy applied through an electric potential.
- #The kinetic energy change is \( \Delta K.E. = \frac{1}{2}mv^2 - \frac{1}{2}m(v_0)^2 \), where \( v_0 \) is the initial velocity, often zero.
It enables the problem-solving mindset needed for developing technologies like renewable energy sources or understanding natural phenomena like lightning. The work-energy principle is more than a mere equation; it's a tool for unlocking how energy influences motion across many contexts.
Other exercises in this chapter
Problem 81
For a d-electron, the orbital angular momentum is (a) \(\sqrt{6 h}\) (b) \(\sqrt{2} \mathrm{~h}\) (c) \(\mathrm{h}\) (d) \(2 \mathrm{~h}\)
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The orbital angular momentum of an electron in \(2 \mathrm{~s}\) orbital is (a) \(+\frac{1}{2} \cdot \frac{\mathrm{h}}{2 \pi}\) (b) zero (c) \(\frac{\mathrm{h}}
View solution Problem 85
A \(600 \mathrm{~W}\) mercury lamp emits monochromatic radiation of wavelength \(331.3 \mathrm{~nm}\). How many photons are emitted from the lamp per second? \(
View solution Problem 88
Correct set of four quantum numbers for the valence (outermost) electron of rubidium \((\mathrm{Z}=37\) ) is (a) \(5,0,0, \pm 1 / 2\) (b) \(6,0,0,+1 / 2\) (c) \
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