/*! 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 The cathode-ray tubes that gener... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 38: Problem 14

The cathode-ray tubes that generated the picture in early color televisions were sources of \(x\) rays. If the acceleration voltage in a television tube is 15.0 \(\mathrm{kV}\) , what are the shortest-wavelength \(\mathrm{x}\) rays produced by the television? (Modern televisions contain shielding to stop these \(x\) rays.)

Short Answer

Expert verified
The shortest wavelength of the x-rays is approximately \(8.26 \times 10^{-11} \text{ m}\).

Step by step solution

01

Understanding the Problem

We are asked to find the shortest-wavelength of the x-rays produced by a television tube when electrons are accelerated through a potential difference of 15.0 kV. This involves using the relationship between energy and wavelength of the x-rays.
02

Using Energy-Wavelength Relation

To find the wavelength, use the equation relating energy and wavelength: \( E = \frac{hc}{\lambda} \). Here, \(E\) is the energy of the x-ray photon, \(h\) is Planck's constant \((6.626 \times 10^{-34} \text{ J s})\), \(c\) is the speed of light \((3.00 \times 10^8 \text{ m/s})\), and \(\lambda\) is the wavelength. Rearranging gives \( \lambda = \frac{hc}{E} \).
03

Calculating the Energy

The energy \(E\) of the x-rays in electron volts (eV) can be calculated from the voltage: \( E = e \times V \), where \(e\) is the elementary charge \((1.602 \times 10^{-19} \text{ C})\), and \(V = 15.0 \text{ kV} = 15,000 \text{ V}\). Therefore, \(E = 1.602 \times 10^{-19} \times 15,000 = 2.403 \times 10^{-15} \text{ J}\).
04

Finding the Shortest Wavelength

Substitute the values into the rearranged formula for wavelength: \(\lambda = \frac{hc}{E}\). This gives \(\lambda = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{2.403 \times 10^{-15}} \approx 8.26 \times 10^{-11} \text{ m}\).
05

Conclusion

Thus, the shortest wavelength of the x-rays produced by the television is approximately \(8.26 \times 10^{-11} \text{ m}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Cathode-Ray Tube
A cathode-ray tube, commonly used in older television sets and computer monitors, is a type of vacuum tube that contains electron guns and a phosphorescent screen. The basic function of a cathode-ray tube is to accelerate and deflect electron beams to create images on the screen. Inside the tube:
  • Electron guns at the back of the tube emit electrons.
  • The electrons are accelerated and focused into a narrow beam.
  • This beam is directed towards a phosphorescent screen at the front of the tube.
  • The screen glows when struck by electrons, creating visible images.
Cathode-ray tubes rely on electric fields to accelerate electrons. When these electrons strike certain materials in the screen, they emit visible light or even x-rays. In early televisions, these x-rays were a by-product of the high-energy electron interactions and required special shielding to prevent exposure.
Energy-Wavelength Relation
In physics, the relationship between the energy of an x-ray and its wavelength is crucial for understanding how imaging and radiation devices work. The core equation that describes this relationship is:\[ E = \frac{hc}{\lambda} \]Where:
  • \(E\) represents the energy of the photon in joules.
  • \(h\) is Planck's constant \(6.626 \times 10^{-34} \text{ J s}\).
  • \(c\) is the speed of light \(3.00 \times 10^8 \text{ m/s}\).
  • \(\lambda\) is the wavelength in meters.
This equation implies that high energy corresponds to short wavelengths. This relationship is exploited in many technological applications, ranging from medical imaging to understanding atomic structures. In the context of x-rays, it is the high energy of the photons that allows them to penetrate materials, making them useful for imaging.
Electron Acceleration Voltage
Electron acceleration voltage is an important concept in the function of cathode-ray tubes and the production of x-rays. It refers to the potential difference through which electrons are accelerated:
  • Electrons are accelerated by an electric field generated by the voltage.
  • The greater the voltage, the higher the energy of the accelerated electrons.
  • Higher energy electrons can result in the emission of x-rays when they hit a material.
In the case of a television tube with an acceleration voltage of 15.0 kV:
  • The electrons gain energy proportional to this voltage, which can be calculated using: \( E = e \times V \).
  • Here, \(e = 1.602 \times 10^{-19} \text{ C}\) (elementary charge).
This energy determines the characteristics of the emitted x-rays, such as their wavelength, with higher acceleration voltages typically producing shorter wavelengths. Such control of x-ray production is vital for both safety and functionality in devices that use these tubes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(\mathrm{X}\) rays with initial wavelength 0.0665 \(\mathrm{nm}\) undergo Compton scattering. What is the longest wavelength found in the scattered \(x\) rays? At which scattering angle is this wavelength observed?

Consider Compton scattering of a photon by a moving electron. Before the collision the photon has wavelength \(\lambda\) and is moving in the \(+x\) -direction, and the electron is moving in the \(-x\) -direction with total energy \(E\) (including its rest energy \(m c^{2} )\) The photon and electron collide head-on. After the collision, both are moving in the \(-x\) -direction (that is, the photon has been scattered by \(180^{\circ}\) . (a) Derive an expression for the wavelength \(\lambda^{\prime}\) of the scattered photon. Show that if \(E>m c^{2},\) where \(m\) is the rest mass of the electron, your result reduces to $$\lambda^{\prime}=\frac{h c}{E}\left(1+\frac{m^{2} c^{4} \lambda}{4 h c E}\right)$$ (b) A beam of infrared radiation from a \(\mathrm{CO}_{2}\) laser \((\lambda=10.6 \mu \mathrm{m})\) collides head-on with a beam of electrons, each of total energy \(E=10.0 \mathrm{GeV}\left(1 \mathrm{GeV}=10^{9} \mathrm{eV}\right) .\) Calculate the wavelength \(\lambda^{\prime}\) of the scattered photons, assuming a \(180^{\circ}\) scattering angle. (c) What kind of scattered photons are these (infrared, microwave, ultraviolet, ett.. \(.\) Can you think of an application of this effect?

A beam of x rays with wavelength 0.0500 nm is Comptonscattered by the electrons in a sample. At what angle from the incident beam should you look to find x rays with a wavelength of (a) \(0.0542 \mathrm{nm} ;\) (b) 0.0521 \(\mathrm{nm}\) ; ( ) 0.0500 \(\mathrm{nm}\) ?

(a) What is the minimum potential difference between the filament and the target of an x-ray tube if the tube is to produce x rays with a wavelength of 0.150 \(\mathrm{nm}\) (b) What is the shortest wavelength produced in an x-ray tube operated at 30.0 \(\mathrm{kV}\) ?

(a) A proton is moving at a speed much slower than the speed of light. It has kinetic energy \(K_{1}\) and momentum \(p_{1} .\) If the momentum of the proton is doubled, so \(p_{2}=2 p_{1},\) how is its new kinetic energy \(K_{2}\) related to \(K_{1} ?\) (b) A photon with energy \(E_{1}\) has momentum \(p_{1} .\) If another photon has momentum \(p_{2}\) that is twice \(p_{1},\) how is the energy \(E_{2}\) of the second photon related to \(E_{1} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Torque and Rotational Motion

Read Explanation

Force

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Solid State Physics

Read Explanation

Classical Mechanics

Read Explanation

Oscillations

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.