Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 36: Problem 33

X-rays of wavelength \(\lambda=0.120 \mathrm{nm}\) are scattered from carbon. What is the Compton wavelength shift for photons detected at \(90.0^{\circ}\) angle relative to the incident beam?

Short Answer

Expert verified
Answer: The Compton wavelength shift for X-rays scattered at a 90-degree angle is 2.43 nm.

Step by step solution

01

Given information and Compton scattering formula

We are given the following information: - Initial wavelength of the X-rays, \(\lambda = 0.120 \, \mathrm{nm}\) - Scattering angle, \(\theta = 90.0^{\circ}\) The Compton scattering formula is given by: $$ \Delta \lambda = \lambda' - \lambda = \frac{h}{m_e c} (1 - \cos{\theta}) $$ Here, \(h\) is the Planck's constant \((6.626 \times 10^{-34} \, \mathrm{J \cdot s})\), \(m_e\) is the electron's mass \((9.109 \times 10^{-31} \, \mathrm{kg})\), \(c\) is the speed of light \((2.998 \times 10^8 \, \mathrm{m/s})\), and \(\Delta \lambda\) is the Compton wavelength shift. We need to find the value of \(\Delta \lambda\).
02

Convert given wavelength to meters

Before we proceed with the calculation, let's convert the given wavelength to meters: $$ \lambda = 0.120 \, \mathrm{nm} \times \frac{1 \, \mathrm{m}}{10^9 \, \mathrm{nm}} = 1.20 \times 10^{-10} \, \mathrm{m} $$
03

Calculate the Compton wavelength using the formula

Now, let's substitute the given values into the Compton scattering formula and compute \(\Delta \lambda\): $$ \Delta \lambda = \frac{h}{m_e c} (1 - \cos{90.0^{\circ}}) $$ $$ \Delta \lambda = \frac{(6.626 \times 10^{-34}\, \mathrm{J\cdot s})}{(9.109 \times 10^{-31} \, \mathrm{kg})(2.998 \times 10^8 \, \mathrm{m/s})}(1 - \cos{90.0^{\circ}}) $$ Since \(\cos{90.0^{\circ}} = 0\), the formula simplifies to: $$ \Delta \lambda = \frac{h}{m_e c} = \frac{(6.626 \times 10^{-34}\, \mathrm{J\cdot s})}{(9.109 \times 10^{-31}\, \mathrm{kg})(2.998 \times 10^8\, \mathrm{m/s})} $$ Calculating the value, we get: $$ \Delta \lambda = 2.43 \times 10^{-12} \, \mathrm{m} $$
04

Convert the Compton wavelength shift to nanometers

Finally, let's convert the Compton wavelength shift to nanometers: $$ \Delta \lambda = 2.43 \times 10^{-12} \, \mathrm{m} \times \frac{10^9 \, \mathrm{nm}}{1 \, \mathrm{m}} = 2.43 \, \mathrm{nm} $$ The Compton wavelength shift for photons detected at a \(90.0^{\circ}\) angle relative to the incident beam is \(\boldsymbol{2.43\, \mathrm{nm}}\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

What is the minimum uncertainty in the velocity of a 1.0 -nanogram particle that is at rest on the head of a 1.0 -mm-wide pin?

What is the wavelength of an electron that is accelerated from rest through a potential difference of \(1.00 \cdot 10^{-5} \mathrm{~V} ?\)

`A pure, defect-free semiconductor material will absorb the electromagnetic radiation incident on that material only if the energy of the individual photons in the incident beam is larger than a threshold value known as the "band-gap" of the semiconductor. The known room-temperature band-gaps for germanium, silicon, and gallium-arsenide, three widely used semiconductors, are \(0.66 \mathrm{eV}, 1.12 \mathrm{eV},\) and \(1.42 \mathrm{eV},\) respectively. a) Determine the room-temperature transparency range of these semiconductors. b) Compare these with the transparency range of \(\mathrm{ZnSe}, \mathrm{a}\) semiconductor with a band-gap of \(2.67 \mathrm{eV},\) and explain the yellow color observed experimentally for the ZnSe crystals. c) Which of these materials could be used for a light detector for the 1550 -nm optical communications wavelength?

The work function of a certain material is \(5.8 \mathrm{eV}\). What is the photoelectric threshold for this material?

A photovoltaic device uses monochromatic light of wavelength 700 . \(\mathrm{nm}\) that is incident normally on a surface of area \(10.0 \mathrm{~cm}^{2}\). Calculate the photon flux rate if the light intensity is \(0.300 \mathrm{~W} / \mathrm{cm}^{2}\).
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Work Energy and Power

Read Explanation

Electricity and Magnetism

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Turning Points in Physics

Read Explanation

Electric Charge Field and Potential

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.

Sign-up for free