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) \(\mathrm{VeV} / \mathrm{m}\) (c) \(\sqrt{2 \mathrm{e}} \mathrm{V} / \mathrm{m}\) (d) none
Step-by-Step Solution
Verified Answer
The velocity acquired by the electron is \(\sqrt{\frac{2eV}{m}}\), which corresponds to option (c).
1Step 1: Understanding the Problem
We have an electron with a charge of \(e\) and mass \(m\). It starts from rest and is accelerated through a potential difference, \(V\). Our task is to find the velocity it will acquire as a result of this acceleration.
2Step 2: Using Energy Conservation
When an electron is accelerated through a potential difference \(V\), the kinetic energy gained is equal to the work done on it by the electric field. The work done is given by \(eV\), and this energy converts to kinetic energy, \(\frac{1}{2}mv^2\), where \(v\) is the velocity.
3Step 3: Setting Up the Equation
According to the conservation of energy, the kinetic energy gained by the electron is equal to the electric potential energy: \(eV = \frac{1}{2}mv^2\).
4Step 4: Solving for Velocity
Rearrange the equation \(eV = \frac{1}{2}mv^2\) to solve for \(v\). Multiply both sides by 2 to get \(2eV = mv^2\), then divide by \(m\) to isolate \(v^2\), resulting in \(v^2 = \frac{2eV}{m}\). Finally, take the square root of both sides to solve for \(v\): \(v = \sqrt{\frac{2eV}{m}}\).
5Step 5: Conclusion and Answer
The expression we derived matches the option (c) \(\sqrt{2eV/m}\). Therefore, the correct choice is option (c).
Key Concepts
Electron AccelerationKinetic Energy and WorkConservation of Energy
Electron Acceleration
When individuals refer to electron acceleration, they are talking about the process that increases the velocity of an electron due to an applied force, usually electric. Initially, an electron at rest will move when subjected to a potential difference, often measured in volts (V). This potential difference creates an electric field that exerts a force on the electron, causing it to accelerate.
Thus, electron acceleration is a key step in various applications such as CRT displays and electron microscopes, where manipulating electron motion is fundamental.
- The electron has a negative charge denoted by \( e \), which leads to a force being applied when in an electric field.
- This force is described by the equation \( F = eE \), where \( E \) is the electric field strength.
- As the electron responds to this force, it begins to accelerate. This is quantitatively described by \( a = \frac{F}{m} = \frac{eE}{m} \), where \( m \) is the electron's mass.
Thus, electron acceleration is a key step in various applications such as CRT displays and electron microscopes, where manipulating electron motion is fundamental.
Kinetic Energy and Work
The relationship between kinetic energy and work is pivotal in understanding how forces cause changes in a particle's motion. Specifically, work is the energy transferred to an object via the movement of another object across a distance by a force. For electrons, when they're accelerated across a potential difference, work is done on them by the electric field.
In a scenario like the one described, the work done on the electron results in it gaining kinetic energy, illustrating the principle that work done on an object results in a change in its kinetic energy.
- The formula for work is \( W = Fd \), where \( F \) is the force applied and \( d \) is the displacement.
- In contexts involving electrons, the work done by the electric field is given by \( W = eV \), where \( e \) is the electron charge and \( V \) is the potential difference.
- Kinetic energy \( KE \) is the energy that an object possesses due to its motion, specifically \( KE = \frac{1}{2}mv^2 \).
In a scenario like the one described, the work done on the electron results in it gaining kinetic energy, illustrating the principle that work done on an object results in a change in its kinetic energy.
Conservation of Energy
The conservation of energy is a fundamental principle stating that within a closed system, energy cannot be created or destroyed; it only transforms from one form to another. This concept is powerfully illustrated in the scenario of an electron accelerating through a potential difference. As the electron moves:
Therefore, the principle of conservation of energy helps us understand that the electron's kinetic energy after acceleration is the result of the work done by the electric field, adhering to the rule that energy is conserved throughout the process.
- The initial electric potential energy is converted into kinetic energy.
- The work-energy principle is applied: the work done by the electric field (\( eV \)) on the electron equals its kinetic energy (\( \frac{1}{2}mv^2 \)).
- This conversion explains that the initial potential energy provided by the electric field becomes the kinetic motion energy of the electron after passing through the potential difference.
Therefore, the principle of conservation of energy helps us understand that the electron's kinetic energy after acceleration is the result of the work done by the electric field, adhering to the rule that energy is conserved throughout the process.
Other exercises in this chapter
Problem 81
For a d-electron, the orbital angular momentum is (a) \(\sqrt{6} \mathrm{~h}\) (b) \(\sqrt{2 h}\) (c) \(\mathrm{h}\) (d) \(2 \mathrm{~h}\)
View solution Problem 82
The orbital angular momentum of an electron in \(2 \mathrm{~s}\) orbital is (a) \(+\frac{1}{2} \cdot \frac{h}{2 \pi}\) (b) zero (c) \(\frac{h}{2 \pi}\) (d) \(\s
View solution Problem 84
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
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? \(
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