/*! 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 31 Crystal diffraction experiments ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1 Ã…, which is of the order of inter-atomic spacing in the lattice) \(\left(m_{e}=9.11 \times 10^{-31} \mathrm{~kg}\right)\).

Short Answer

Expert verified
X-rays have greater energy than electrons for the same wavelength.

Step by step solution

01

Understand the Problem

To compare the energies of X-rays and electrons with the same wavelength, we need to use the relationship between energy, wavelength, and momentum in different forms: for X-rays, the energy is in terms of frequency, and for electrons, it's related to kinetic energy.
02

Energy of X-rays

The energy of X-rays is given by the equation:\[ E = \frac{hc}{\lambda} \]where \( h = 6.626 \times 10^{-34} \text{ J s} \) is Planck's constant, \( c = 3.00 \times 10^8 \text{ m/s} \) is the speed of light, and \( \lambda = 1 \times 10^{-10} \text{ m} \) is the given wavelength. By substituting these values, we find the energy of the X-rays.
03

Calculate Energy of X-rays

Substitute the values into the equation:\[ E = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{1 \times 10^{-10}} = 1.99 \times 10^{-15} \text{ J} \]This is the energy of the X-rays at the given wavelength.
04

Energy of Electrons (De Broglie Wavelength)

For electrons, we use the de Broglie wavelength formula:\[ \lambda = \frac{h}{p} \]where \( p \) is momentum. The kinetic energy \( K \) of electrons is given by:\[ K = \frac{p^2}{2m} \]We solve for \( p \) in terms of \( \lambda \) and substitute it into the kinetic energy formula.
05

Solve for Electron Momentum

Rearranging for \( p \) in terms of \( \lambda \):\[ p = \frac{h}{\lambda} \]Thus:\[ p^2 = \left(\frac{h}{\lambda}\right)^2 \]
06

Calculate Kinetic Energy of Electrons

Substitute \( p^2 \) into the kinetic energy formula:\[ K = \frac{h^2}{2m \lambda^2} \]Substitute the known values:\[ K = \frac{(6.626 \times 10^{-34})^2}{2 \times 9.11 \times 10^{-31} \times (1 \times 10^{-10})^2} = 1.20 \times 10^{-17} \text{ J} \]This is the kinetic energy of the electron.
07

Compare Energies

Compare the two energies calculated. The energy of the X-ray is \( 1.99 \times 10^{-15} \text{ J} \) and the energy of the electron is \( 1.20 \times 10^{-17} \text{ J} \). The X-rays have higher energy.

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 91Ó°ÊÓ!

Key Concepts

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

X-ray energy
X-ray energy is an essential aspect when exploring crystal diffraction experiments. X-rays are a form of electromagnetic radiation, similar to light but with much shorter wavelengths. The energy of X-rays correlates with their frequency and is determined using Planck's constant.

The energy can be calculated using the formula:
  • \( E = \frac{hc}{\lambda} \)
where:
  • \( h \) is Planck's constant, valued at approximately \( 6.626 \times 10^{-34} \text{ J s} \).
  • \( c \) is the speed of light, approximated as \( 3.00 \times 10^8 \text{ m/s} \).
  • \( \lambda \) represents the wavelength, typically around \( 1 \times 10^{-10} \text{ m} \) for this problem.
For crystal diffraction analysis, the precise estimation of X-ray energy assists in mapping out how X-rays interact with atomic structures. Shorter wavelengths, like those of X-rays, offer high energy, making them suitable for probing tiny distances between atoms in a crystal lattice.

Hence, knowing the precise energy helps understand X-ray diffraction patterns, critical for analyzing crystal structures.
Electron kinetic energy
Kinetic energy in electrons is pivotal when considering electron diffraction. Unlike X-rays, electrons are particles with mass, which influences their behavior and characteristics. As they are accelerated through a potential difference, they gain kinetic energy.

The kinetic energy of electrons can be represented using their momentum:
  • \( K = \frac{p^2}{2m} \)
where:
  • \( p \) symbolizes the momentum of the electron, defined by \( p = \frac{h}{\lambda} \).
  • \( m \) is the electron mass, approximately \( 9.11 \times 10^{-31} \text{ kg} \).
The calculation of the kinetic energy reveals the electrons' ability to interact with a crystal lattice.

Their comparative lower energy than that of X-rays means electron diffraction can offer enhanced resolution for certain types of material investigations but usually involves different setups or interpretations of diffraction patterns.
De Broglie wavelength
The De Broglie wavelength is a core concept in quantum mechanics, significant for understanding particle-wave duality. It indicates that particles such as electrons exhibit wave-like properties, like diffraction and interference, typically observed in waves.

The wavelength is calculated using De Broglie's formula:
  • \( \lambda = \frac{h}{p} \)
where:
  • \( \lambda \) is the wavelength associated with a particle.
  • \( h \) is Planck's constant.
  • \( p \) denotes the particle's momentum.
For electrons, this wavelength helps determine how they will diffract through a crystal.

By associating wave-like characteristics with particles, De Broglie's hypothesis bridges classical and quantum physics, providing a framework for predicting how particles behave as waves, notably when their energies and wavelengths mirror those used in X-ray diffraction experiments.

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

Answer the following questions: (a) Quarks inside protons and neutrons are thought to carry fractional charges \([(+2 / 3) e ;(-1 / 3) e] .\) Why do they not show up in Millikan's oil-drop experiment? (b) What is so special about the combination \(e / m ?\) Why do we not simply talk of \(e\) and \(m\) separately? (c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures? (d) Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic? Why is there an energy distribution of photoelectrons? (e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations: \(E=h v, p=\frac{h}{\lambda}\) But while the value of \(\lambda\) is physically significant, the value of \(v\) (and therefore, the value of the phase speed \(v \lambda\) ) has no physical significance. Why?

Light of intensity \(10^{-5} \mathrm{~W} \mathrm{~m}^{-2}\) falls on a sodium photo-cell of surface area \(2 \mathrm{~cm}^{2}\). Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about \(2 \mathrm{eV}\). What is the implication of your answer?

The photoelectric cut-off voltage in a certain experiment is \(1.5 \mathrm{~V}\). What is the maximum kinetic energy of photoelectrons emitted?

An electron microscope uses electrons accelerated by a voltage of \(50 \mathrm{kV}\). Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?

Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).

See all solutions

Recommended explanations on Physics Textbooks

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.