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An electron is accelerated from rest through a potential difference of \(2.50 \mathrm{MV}\). Find the electron's (a) speed, (b) kinetic energy, and (c) momentum.

Short Answer

Expert verified
The speed is approximately \(2.98 \times 10^8 \, \text{m/s}\), kinetic energy is \(4.005 \times 10^{-13} \, \text{J}\), and momentum is \(2.715 \times 10^{-22} \, \text{kg m/s}\).

Step by step solution

01

Calculate the Kinetic Energy

First, find the kinetic energy the electron gains as it's accelerated. The electric potential energy is converted into kinetic energy as the electron is accelerated through a potential difference. The formula for kinetic energy when a charge is accelerated through a potential difference is given by:\[K = qV\]where \(q\) is the charge of the electron \((1.602 \times 10^{-19} \, \text{C})\) and \(V\) is the potential difference \((2.50 \times 10^6 \, \text{V})\). Substituting the values gives:\[K = (1.602 \times 10^{-19} \, \text{C})(2.50 \times 10^6 \, \text{V}) = 4.005 \times 10^{-13} \, \text{J}\]
02

Calculate the Speed

To find the speed of the electron, use the relation between kinetic energy and speed:\[K = \frac{1}{2}mv^2\]where \(m\) is the mass of the electron \((9.109 \times 10^{-31} \, \text{kg})\) and \(v\) is the speed. Solving for \(v\) gives:\[v = \sqrt{\frac{2K}{m}}\]Substitute \(K = 4.005 \times 10^{-13} \, \text{J}\) and \(m = 9.109 \times 10^{-31} \, \text{kg}\):\[v = \sqrt{\frac{2 \times 4.005 \times 10^{-13} \, \text{J}}{9.109 \times 10^{-31} \, \text{kg}}} \approx 2.98 \times 10^8 \, \text{m/s}\]
03

Calculate the Momentum

The momentum \(p\) of the electron can be found using the relation:\[p = mv\]Now, substitute the calculated speed \(v = 2.98 \times 10^8 \, \text{m/s}\) and the mass \(m = 9.109 \times 10^{-31} \, \text{kg}\):\[p = (9.109 \times 10^{-31} \, \text{kg})(2.98 \times 10^8 \, \text{m/s}) \approx 2.715 \times 10^{-22} \, \text{kg m/s}\]

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy that an object has due to its motion. It plays a central role in understanding how an electron behaves when it is accelerated by a potential difference. The formula to calculate kinetic energy, when an electric charge is accelerated, is \[ K = qV \]where \( q \) stands for electric charge, and \( V \) refers to the potential difference. For an electron, which always carries a charge of \( 1.602 \times 10^{-19} \) Coulombs, this relationship provides a direct path to calculate kinetic energy under an applied voltage, or potential difference.
  • Kinetic energy is directly proportional to potential difference.
  • In the exercise, the kinetic energy gained by the electron equals \( 4.005 \times 10^{-13} \) joules.
In this context, every joule represents how the electron's motion ramps up as it speeds through the electric field.
Potential Difference
Potential difference, often measured in volts, is what drives the movement of electrons and governs their acceleration in electronic systems. It is the electrical potential energy difference between two points in an electric field and is vital for calculating the kinetic energy of a charged particle.
  • Expressed in units of volts, where \(1 \) volt equals \( 1 \) joule per Coulomb.
  • The exercise example used 2.50 MV, meaning 2.50 million volts make the electron speed up.
For an electron, crossing through a potential difference of this magnitude is comparable to launching from a standing position to high speeds in mere nanoseconds. The greater the potential difference, the more energy is imparted, pushing the electron along its path with a significant force.
Electric Charge
Electric charge is a fundamental property of matter, essentially manifesting in positive and negative forms. Electrons, carrying a negative charge, engage in various interactions within an electric field that determines their movement.
  • Electron charge: \(-1.602 \times 10^{-19}\) Coulombs, a constant value and a cornerstone of particle physics.
  • In calculations, like the exercise, this charge is crucial for determining energy transformations in fields.
Despite its supposition as negative, in current conventions, electron flow is depicted in opposite direction to current's flow. Hence, the negative quality pertinent to electrons accentuates not only in equations like \(K = qV\), but also in understanding electron pathways through circuits.
Electron Speed
Electron speed after being accelerated by an electric field is critical in evaluating energy transfer in electronic devices. When seeking the electron's final speed, kinetic energy formulas become invaluable.
  • The derived formula \( v = \sqrt{\frac{2K}{m}} \) helps compute speed once kinetic energy \( K \) is determined.
  • For example, the homework solution resulted in a speed of approximately \( 2.98 \times 10^8 \) m/s.
This calculation illustrates how an electron can reach near-light speed for high potential differences. Fast-moving electrons become essential in solid-state devices, where speed impacts energy conduction and bandwidth in semiconductors. The interplay of speed, energy, and potential curates an intricate dance of physics that underpins modern technology.

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Most popular questions from this chapter

Alpha Centauri, a star close to our solar system, is about 4.3 light-years away. Suppose a spaceship traveled this distance at a constant speed of \(0.90 c\) relative to Earth. (a) How long did the trip take according to an Earth- based clock? (b) How long did the trip take according to the traveler's clock? Which of you answers to parts (a) and (b) are proper time intervals, if any? (c) What is the trip distance according to an Earth-based observer? (d) What is the trip distance according to the traveler? Which of your answers to parts (b) and (c) are proper lengths, if any?

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