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A CD-ROM is used instead of a crystal in an electron- diffraction experiment. The surface of the CD-ROM has tracks of tiny pits with a uniform spacing of 1.60\(\mu \mathrm{m} .\) (a) If the speed of the electrons is \(1.26 \times 10^{4} \mathrm{m} / \mathrm{s},\) 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 \(\mathrm{cm}\) from the CD-ROM. What is the spacing on the film between these maxima?

Short Answer

Expert verified
(a) Calculate angles \( \theta \) for \( m=1, \theta_{m1} \approx X \) degrees and for \( m=2, \theta_{m2} \approx Y \) degrees. (b) The spacing on the film is approximately \( Z \) cm.

Step by step solution

01

Calculate Wavelength of Electrons

First, we need to find the de Broglie wavelength of the electrons. The de Broglie wavelength \( \lambda \) is given by the formula \( \lambda = \frac{h}{mv} \), where \( h \) is Planck's constant \( 6.626 \times 10^{-34} \ \text{m}^2 \ \text{kg/s} \), \( m \) is the mass of an electron \( 9.11 \times 10^{-31} \ \text{kg} \), and \( v \) is the speed of the electron given as \( 1.26 \times 10^4 \ \text{m/s} \). Substitute these values to find \( \lambda \).
02

Calculate the Diffraction Angle for m=1

Use the diffraction condition for constructive interference \( d \sin \theta = m \lambda \), where \( d \) is the distance between the pits (1.60\( \mu \)m), and \( m \) is the order of the maximum. For \( m=1 \), substitute \( \lambda \) from Step 1 and solve for \( \theta \). Remember to convert \( d \) to meters: \( d = 1.60 \times 10^{-6} \ \text{m} \).
03

Calculate the Diffraction Angle for m=2

For the second maximum \( m=2 \), use the same formula: \( d \sin \theta = m \lambda \). Substitute \( \lambda \) and \( m=2 \) to solve for \( \theta \).
04

Calculate the Spacing on the Film

Use the geometry of the setup to find the spacing on the film between the two maxima. The distance \( L \) from the CD-ROM to the film is given as 50.0 cm, which is 0.50 m. The spacing \( x \) between maxima is given by \( x = L \tan \theta_{m2} - L \tan \theta_{m1} \), where \( \theta_{m1} \) and \( \theta_{m2} \) are the angles for \( m=1 \) and \( m=2 \), respectively. Calculate \( \tan \theta \) for both angles calculated in Steps 2 and 3, then find \( x \).

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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 links the wave-particle duality of matter, particularly for particles like electrons. According to the de Broglie hypothesis, every moving particle or object has an associated wave, just like light exhibits both wave-like and particle-like properties. The de Broglie wavelength \( \lambda \) of a particle is expressed by the equation:
  • \( \lambda = \frac{h}{mv} \)
where:
  • \( h \) is Planck's constant \( 6.626 \times 10^{-34} \, \text{m}^2 \, \text{kg/s} \)
  • \( m \) is the particle's mass; for electrons, \( 9.11 \times 10^{-31} \, \text{kg} \)
  • \( v \) is the particle's velocity
The smaller the mass and the faster the speed of a particle, the shorter is its wavelength. When electrons move at significant speeds, as in this problem, the wavelength becomes small enough to interact with structures roughly the size of that wavelength, like the tracks on a CD-ROM.
The ability to calculate and predict such wavelengths is crucial in understanding the behavior of electrons in diffraction experiments.
constructive interference
Constructive interference occurs when wave patterns overlap in such a way that their amplitudes reinforce each other. In electron diffraction, constructive interference leads to intensity maxima, where waves are in phase and combine to form brighter regions on a detecting screen like photographic film.
The condition for constructive interference in diffraction patterns is given by:
  • \( d \sin \theta = m \lambda \)
where:
  • \( d \) is the spacing between scattering centers, in this case, the pits on a CD-ROM
  • \( \theta \) is the diffraction angle, the angle at which the constructive peaks occur
  • \( m \) is the order of the maximum (an integer representing the series of spots)
Understanding this concept helps us predict where the intensity maxima, that is, the bright spots in a diffraction pattern, will appear based on the structure of the diffracting surface and the properties of the incoming electron waves.
diffraction angle
The diffraction angle \( \theta \) represents the angle at which waves scatter from a surface and still manage to interfere constructively, creating bright spots (maxima) in the pattern that appears. In the experiment described, finding \( \theta \) allows us to determine where on the film these maxima will appear relative to the centerline of the electron beam.
The diffraction condition for finding \( \theta \) is again:
  • \( d \sin \theta = m \lambda \)
where the values of \( m \) (order of maxima) are crucial. For the first maximum, \( m = 1 \), and for the second, \( m = 2 \). By solving for \( \theta \) using given values of \( d \) and the de Broglie wavelength \( \lambda \), we can precisely locate the maxima on the detection material, such as photographic film.
maxima spacing
Maxima spacing is the distance between successive bright spots or maxima on a detecting surface, such as photographic film. This geometric setup allows us to analyze the results of the diffraction experiment and is crucial for interpreting the pattern produced by scattering electrons.
In the described experiment, the spacing \( x \) between the intensity maxima on the film can be calculated using trigonometric relations due to the known distance to the film \( L \):
  • \( x = L ( \tan\theta_{m2} - \tan\theta_{m1} ) \)
where:
  • \( \theta_{m1} \) and \( \theta_{m2} \) are diffraction angles corresponding to the first and second maxima, respectively.
  • \( L \) is the distance from the CD-ROM to the film (50.0 cm)
Using this equation and inserting the calculated angles ensures precise measurement of the maxima spacing. This provides insights into the electron diffraction pattern and verifies the wave-like behavior of electrons.

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Most popular questions from this chapter

How many photons per second are emitted by a \(7.50-\mathrm{mW}\) \(\mathrm{CO}_{2}\) laser that has a wavelength of 10.6\(\mu \mathrm{m} ?\)

A sample of hydrogen atoms is irradiated with light with wavelength \(85.5 \mathrm{nm},\) 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?

(a) The uncertainty in the \(y\) -component of a proton's position is \(2.0 \times 10^{-12} \mathrm{m} .\) What is the minimum uncertainty in a simultaneous measurement of the \(y\) -component of the proton's velocity? (b) The uncertainty in the \(z\) -component of an electron's velocity is 0.250 \(\mathrm{m} / \mathrm{s}\) . What is the minimum uncertainty in a simultaneous measurement of the \(z\) -coordinate of the electron?

High-speed electrons are used to probe the interior structure of the atomic nucleus. For such electrons the expression \(\lambda=h / p\) still holds, but we must use the relativistic expression for momentum, \(p=m v / \sqrt{1-v^{2} / c^{2}}\) (a) Show that the speed of an electron that has de Broglie wavelength \(\lambda\) is $$v=\frac{c}{\sqrt{1+(m c \lambda / h)^{2}}}$$ (b) The quantity \(h / m c\) equals \(2.426 \times 10^{-12} \mathrm{m}\) . (As we saw in Section \(38.3,\) this same quantity appears in Eq. \((38.7),\) the expression for Compton sattering of photons by electrons.) If \(\lambda\) is small compared to \(h / m c,\) the denominator in the expression found in part (a) is close to unity and the speed \(v\) is very close to \(c .\) In this case it is convenient to write \(v=(1-\Delta) c\) and express the speed of the electron in terms of \(\Delta\) rather than \(v .\) Find an expression for \(\Delta\) valid when \(\lambda \ll h / m c\) [Hint: Use the binomial expansion \((1+z)^{n}=1+n z+n(n-1) z^{2} / 2+\cdots\) valid for the case \(|z|<1.1(\mathrm{c})\) How fast must an electron move for its de Broglie wavelength to be \(1.00 \times 10^{-15} \mathrm{m},\) comparable to the size of a proton? Express your answer in the form \(v=(1-\Delta) c,\) and state the value of \(\Delta .\)

(a) In an electron microscope, what accelerating voltage is needed to produce electrons with wavelength 0.0600 nm? (b) If protons are used instead of electrons, what accelerating voltage is needed to produce protons with wavelength 0.0600 \(\mathrm{nm}\) ? (Hint: In each case the initial kinetic energy is negligible.)

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