/*! 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 25 Cathode-ray tube. A cathode-ray ... [FREE SOLUTION] | 91Ó°ÊÓ

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Cathode-ray tube. A cathode-ray tube (CRT) is an evacuated glass tube. Electrons are produced at one end, usually by the heating of a metal. After being focused electromagnetically into a beam, they are accelerated through a potential difference, called the accelerating potential. The electrons then strike a coated screen, where they transfer their energy to the coating through collisions, causing it to glow. CRTs are found in oscilloscopes and computer monitors, as well as in earlier versions of television screens. (a) If an electron of mass \(m\) and charge \(-e\) is accelerated from rest through an accelerating potential \(V,\) show that the speed it gains is \(v=\sqrt{2 e V / m}\) . We are assuming that \(V\) is small enough that the final speed is much less than the speed of light.) (b) If the accelerating potential is \(95 \mathrm{V},\) how fast will the electrons be moving when they hit the screen?

Short Answer

Expert verified
(a) The speed is \( v = \sqrt{\frac{2eV}{m}} \). (b) Electrons move at \( 5.8 \times 10^6 \text{ m/s} \).

Step by step solution

01

Understanding the Energy Conversion

When an electron is initially at rest, and it's accelerated through a potential difference \( V \), the work done on it is converted into kinetic energy. The work done, due to the electric field, equals the charge \( -e \) times the potential difference \( V \), which is \( eV \) since we consider just the magnitude here. This energy goes into accelerating the electron.
02

Setting Up the Kinetic Energy Equation

The kinetic energy \( KE \) of an electron with mass \( m \) and velocity \( v \) is given by \( KE = \frac{1}{2}mv^2 \). Since the work done on the electron is equal to its kinetic energy, we set \( \frac{1}{2}mv^2 = eV \).
03

Solving for Velocity

Rearrange the equation \( \frac{1}{2}mv^2 = eV \) to express \( v \). Multiply both sides by 2, giving \( mv^2 = 2eV \). Then, divide both sides by \( m \) to get \( v^2 = \frac{2eV}{m} \). Finally, take the square root of both sides to solve for \( v \): \( v = \sqrt{\frac{2eV}{m}} \). This concludes part (a) of the problem.
04

Calculating the Electron Speed for Part (b)

Given \( V = 95 \mathrm{V} \), we can now calculate the speed. We use \( v = \sqrt{\frac{2eV}{m}} \). Substitute \( e \approx 1.6 \times 10^{-19} \mathrm{C} \) (the charge of an electron) and \( m \approx 9.11 \times 10^{-31} \mathrm{kg} \) (the mass of an electron): \( v = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 95}{9.11 \times 10^{-31}}} \). Calculate this to find \( v \approx 5.8 \times 10^6 \text{ m/s} \).

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Key Concepts

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

Electron Acceleration
In a cathode-ray tube, electron acceleration is a crucial step that transforms the electron's potential energy into kinetic energy. Electrons are released from the cathode and begin with essentially no speed. By applying a potential difference, these electrons are accelerated.
The acceleration is due to the electric field created by the potential difference across the CRT. Electrons, which are negatively charged particles, are propelled towards the positively charged anode, gaining speed rapidly.
This acceleration is akin to rolling a ball downhill, where the potential energy is converted into kinetic energy, increasing the ball's speed as it descends.
Kinetic Energy
Kinetic energy is the energy which a particle possesses due to its motion. For electrons in a cathode-ray tube, this energy is transferred from the work done on the electron by the electric field. The formula for kinetic energy is expressed as \( KE = \frac{1}{2}mv^2 \), where \( m \) is mass and \( v \) is velocity.
When an electron is accelerated and starts moving faster, it gains kinetic energy. This gain in energy comes from the potential energy sourced from the potential difference in the CRT.
Electrons eventually hit the screen of the cathode-ray tube, transferring kinetic energy into light energy, causing a visible glow on the screen.
Potential Difference
The potential difference, often referred to as voltage, is a measure of the difference in electric potential energy per unit charge between two points. In the context of a cathode-ray tube, this potential difference is the driving force that accelerates the electrons.
The electric potential difference is fundamental in determining how much kinetic energy an electron will gain. It is directly proportional to the work done on the electron: \( W = eV \). This shows that a higher potential difference results in greater acceleration for the electrons.
The units of potential difference are volts (V), and even a relatively small potential, like 95 volts, is enough to give significant speed to electrons.
Electric Field
The electric field in a cathode-ray tube is generated by the potential difference between the cathode and the anode. This field is what propels the electrons forward, acting as a force that moves them from areas of low potential energy to areas of high potential energy.
The strength of the electric field can be given by the rate of change of the electric potential with distance between these two electrodes. It is this field that plays a pivotal role in converting the potential energy into kinetic energy, facilitating the motion of electrons.
Understanding the behavior of electric fields is essential, as it helps in grasping how various devices, including CRTs, are able to manipulate electron flow for different applications.

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Most popular questions from this chapter

You are working on an electronics project requiring a variety of capacitors, but have only a large supply of 100 nF capacitors available. Show how you can connect these capacitors to produce each of the following equivalent capacitances: (a) \(50 \mathrm{nF},\) (b) \(450 \mathrm{nF},(\mathrm{c}) 25 \mathrm{nF},\) (d) 75 \(\mathrm{nF.}\)

(a) An electron is to be accelerated from \(3.00 \times 10^{6} \mathrm{m} / \mathrm{s}\right.\) to \(8.00 \times 10^{6} \mathrm{m} / \mathrm{s}\) . Through what potential difference must the electron pass to accomplish this? (b) Through what potential difference must the electron pass if it is to be slowed from \(8.00 \times 10^{6} \mathrm{m} / \mathrm{s}\) to a halt?

In a particular Millikan oil-drop apparatus, the plates are 2.25 \(\mathrm{cm}\) apart. The oil used has a density of \(0.820 \mathrm{g} / \mathrm{cm}^{3},\) and the atomizer that sprays the oil drops produces drops of diameter \(1.00 \times 10^{-3} \mathrm{mm}\) . (a) What strength of electric field is needed to hold such a drop stationary against gravity if the drop contains five excess electrons? (b) What should be the potential difference across the plates to produce this electric field? (c) If another drop of the same oil requires a plate potential of 73.8 \(\mathrm{V}\) to hold it stationary, how many excess electrons did it contain?

A 10.0\(\mu \mathrm{F}\) parallel-plate capacitor with circular plates is connected to a 12.0 \(\mathrm{V}\) battery. (a) What is the charge on each plate? (b) How much charge would be on the plates if their separation were doubled while the capacitor remained connected to the battery? (c) How much charge would be on the plates if the capacitor were connected to the 12.0 \(\mathrm{V}\) battery after the radius of each plate was doubled without changing their separation?

A point charge \(q_{1}=+2.40 \mu C\) is held stationary at the origin. A second point charge \(q_{2}=-4.30 \mu C\) moves from the point \(x=0.150 \mathrm{m}, y=0,\) to the point \(x=0.250 \mathrm{m},\) \(y=0.250 \mathrm{m} .\) How much work is done by the electric forceon \(q_{2} ?\)

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