Question

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

(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

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: \[\lambda = \frac{6.626 \times 10^{-34}}{(9.11 \times 10^{-31})(1.26 \times 10^{4})}\]

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.

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 \(m = 2\): \(1.60 \times 10^{-6} \sin \theta = 2 \times 5.79 \times 10^{-8}\)

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:- \(\theta_{m=1} \approx 2.08^\circ\) - \(\theta_{m=2} \approx 4.15^\circ\)

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.

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, \(\Delta y \approx 0.018 \text{ m} = 1.8 \text{ cm}\).

Key concepts

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.