Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A CD-ROM is used instead of a crystal in an electrondiffraction experiment. The surface of the CD-ROM has tracks of tiny pits with a uniform spacing of 1.60 \(\mu{m}\). (a) If the speed of the electrons is 1.26 \(\times\) 10\(^4\) m/s, at which values of \(\theta\) will the \(m\) = 1 and \(m\) = 2 intensity maxima appear? (b) The scattered electrons in these maxima strike at normal incidence a piece of photographic film that is 50.0 cm from the CD-ROM. What is the spacing on the film between these maxima?

Short Answer

Expert verified
(a) \(\theta_{m=1} \approx 2.08^\circ\), \(\theta_{m=2} \approx 4.15^\circ\). (b) Spacing \(\approx 1.8 \text{ cm}\).

Step by step solution

01

Understanding Electron Wavelength

First, we need to determine the de Broglie wavelength of the electrons, which is given by \(\lambda = \frac{h}{mv}\), where \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)), \(m\) is the electron mass (approximated as \(9.11 \times 10^{-31} \, \text{kg}\)), and \(v\) is the speed of the electrons. Substitute the given speed into the equation: \[\lambda = \frac{6.626 \times 10^{-34}}{(9.11 \times 10^{-31})(1.26 \times 10^{4})}\]
02

Calculate De Broglie Wavelength

After substituting the values into the wavelength equation, we find: \(\lambda \approx 5.79 \times 10^{-8} \, \text{m}\). This is the wavelength of the electrons used in this diffraction experiment.
03

Determine Angles for Maxima

We use the equation for constructive interference: \(d \sin \theta = m \lambda\), where \(d = 1.60 \times 10^{-6} \, \text{m}\) and \(\lambda = 5.79 \times 10^{-8} \, \text{m}\). For \(m = 1\) and \(m = 2\), separately solve for \(\theta\): - For \(m = 1\): \(1.60 \times 10^{-6} \sin \theta = 5.79 \times 10^{-8}\) - For \(m = 2\): \(1.60 \times 10^{-6} \sin \theta = 2 \times 5.79 \times 10^{-8}\)
04

Solve for \\(\theta\\)

Using the relationship from the previous step:- For \(m = 1\): \(\sin \theta = \frac{5.79 \times 10^{-8}}{1.60 \times 10^{-6}} \approx 0.0362\).- For \(m = 2\): \(\sin \theta = \frac{2 \times 5.79 \times 10^{-8}}{1.60 \times 10^{-6}} \approx 0.0724\).Using inverse sine, we find:- \(\theta_{m=1} \approx 2.08^\circ\) - \(\theta_{m=2} \approx 4.15^\circ\)
05

Calculate Spacing on Film

Use the formula for spacing \(y = L \tan \theta\), where \(L = 0.50 \text{ m}\). Calculate for \(\theta = 2.08^\circ\) and \(\theta = 4.15^\circ\):- \(y_{m=1} = 0.50 \times \tan(2.08^\circ)\) - \(y_{m=2} = 0.50 \times \tan(4.15^\circ)\)Find the difference in spacing between these maxima.
06

Calculate Film Separation

Convert angles to radians and compute:- \(y_{m=1} = 0.50 \times \tan(2.08^\circ \approx 0.0363)\) - \(y_{m=2} = 0.50 \times \tan(4.15^\circ \approx 0.0726)\)Find the spacing: \(\Delta y = y_{m=2} - y_{m=1}\). After calculation, \(\Delta y \approx 0.018 \text{ m} = 1.8 \text{ cm}\).

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 Vaia!

Key Concepts

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

de Broglie Wavelength
The concept of de Broglie wavelength is essential for understanding how particles like electrons can exhibit wavelike behavior. According to the de Broglie hypothesis, every moving particle or object has an associated wavelength, known as its de Broglie wavelength.
  • This wavelength is calculated using the formula: \( \lambda = \frac{h}{mv} \).
  • \(h\) represents Planck's constant, a fundamental quantity in quantum mechanics known to be approximately \(6.626 \times 10^{-34} \, \text{Js}\).
  • \(m\) refers to the mass of the particle, in this case, the electron, which has a mass of about \(9.11 \times 10^{-31} \, \text{kg}\).
  • \(v\) is the velocity of the particle.
For this exercise, the velocity of electrons is given, and using the equation above, we can calculate their de Broglie wavelength to be \(\approx 5.79 \times 10^{-8} \, \text{m}\). This wavelength is pivotal for the calculations in the diffraction process, providing the basis for understanding how electrons interact with a CD-ROM surface similarly to light waves.
Constructive Interference
Constructive interference occurs when two or more waves overlap to produce a wave with a larger amplitude. In electron diffraction, it defines the conditions under which electrons will show enhanced intensity on a detection screen. This results in regions of maximum brightness, also called intensity maxima.The condition for constructive interference in a diffraction pattern is expressed as:
  • \(d \sin \theta = m \lambda\),
where
  • \(d\) is the distance between diffraction lines, in this exercise, it's the pit spacing on the CD-ROM, \(1.60 \times 10^{-6} \, \text{m}\).
  • \(m\) is the order of the maxima, which can be any integer (1 for the first maxima, 2 for the second, etc.).
  • \(\lambda\) is the de Broglie wavelength of the electrons.
  • \(\theta\) is the angle at which these maxima occur.
By substituting the known values into this equation for \(m = 1\) and \(m = 2\), we calculate the angles \(\theta\) for the first and second maxima of intensity.
Angle of Diffraction
The angle of diffraction is a key factor in determining where the interference patterns will appear in an experiment setup, like in the case of the CD-ROM used as a diffraction grating in this problem.The equation \(d \sin \theta = m \lambda\) helps calculate the specific angles \(\theta\) where we observe the intensity maxima for given orders.For this specific exercise:
  • For \(m = 1\), the angle is calculated as \(\theta_{m=1} \approx 2.08^\circ\).
  • For \(m = 2\), \(\theta_{m=2} \approx 4.15^\circ\).
These angles signify the directions at which the scattered electrons constructively interfere to produce visible maxima on a film placed at a certain distance. Calculating these angles accurately is vital for predicting where these electric patterns can be visualized.
Electron Velocity
Understanding electron velocity is fundamental to solving this exercise because it's one of the parameters affecting the de Broglie wavelength.The velocity of electrons in this problem is given as \(1.26 \times 10^4 \, \text{m/s}\), representing how fast these electrons are traveling in the experimental setup. Velocity affects not just the wavelength, as observed in the de Broglie relationship \( \lambda = \frac{h}{mv} \), but also influences:
  • The energy of the electrons, which is proportional to the square of the velocity.
  • The resolution of the diffraction pattern, as changes in speed can shift the positions of interference maxima.
Thus, accurately incorporating electron velocity ensures precise calculations of other parameters, such as the angles at which constructive interference occurs and the resulting spacing between maxima on photographic films in experimental setups.

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

The negative muon has a charge equal to that of an electron but a mass that is 207 times as great. Consider a hydrogenlike atom consisting of a proton and a muon. (a) What is the reduced mass of the atom? (b) What is the ground-level energy (in electron volts)? (c) What is the wavelength of the radiation emitted in the transition from the \(n\) = 2 level to the \(n\) = 1 level?

A beam of electrons is accelerated from rest through a potential difference of 0.100 kV and then passes through a thin slit. When viewed far from the slit, the diffracted beam shows its first diffraction minima at \(\pm\)14.6\(^\circ\) from the original direction of the beam. (a) Do we need to use relativity formulas? How do you know? (b) How wide is the slit?

A scientist has devised a new method of isolating individual particles. He claims that this method enables him to detect simultaneously the position of a particle along an axis with a standard deviation of 0.12 nm and its momentum component along this axis with a standard deviation of 3.0 \(\times\) 10\(^{-25}\) kg \(\bullet\) m/s. Use the Heisenberg uncertainty principle to evaluate the validity of this claim.

A sample of hydrogen atoms is irradiated with light with wavelength 85.5 nm, and electrons are observed leaving the gas. (a) If each hydrogen atom were initially in its ground level, what would be the maximum kinetic energy in electron volts of these photoelectrons? (b) A few electrons are detected with energies as much as 10.2 eV greater than the maximum kinetic energy calculated in part (a). How can this be?

If your wavelength were 1.0 m, you would undergo considerable diffraction in moving through a doorway. (a) What must your speed be for you to have this wavelength? (Assume that your mass is 60.0 kg.) (b) At the speed calculated in part (a), how many years would it take you to move 0.80 m (one step)? Will you notice diffraction effects as you walk through doorways?

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.

Sign-up for free