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CP A beam of electrons is accelerated from rest and then passes through a pair of identical thin slits that are \(1.25 \mathrm{nm}\) apart. You observe that the first double-slit interference dark fringe occurs at \(\pm 18.0^{\circ}\) from the original direction of the beam when viewed on a distant screen. (a) Are these electrons relativistic? How do you know? (b) Through what potential difference were the electrons accelerated?

Short Answer

Expert verified
The electrons are not relativistic because their speed is much less than the speed of light. The potential difference through which the electrons were accelerated can be calculated using the kinetic energy formula and will be your final answer.

Step by step solution

01

Calculating the speed of electrons

To determine if the electrons are relativistic, we first need to calculate their speed. We can use the double-slit interference formula in quantum mechanics: \( \theta = \arcsin(\frac{m \lambda v}{d}) \). Let's rearrange it for speed, solving for v: \( v = \frac{d \sin(\theta)}{m \lambda} \). Given that the first dark fringe is at \(\pm 18.0^{\circ}\) (\( \theta = 18.0^{\circ}\)), the slit separation is \(1.25 \times 10^{-9}\ m\) (d), the mass of electron \(m = 9.11 \times 10^{-31}\ Kg\) and the wavelength of electron \(\lambda = 2.43 \times 10^{-12}\ m\) (which is the de Broglie wavelength), we plug these values into the formula and calculate v.
02

Checking if electrons are relativistic

The speed of light is approximately \(3.00 \times 10^8\ m/s\). If an electron's speed is close to this, then it is relativistic. After calculating from step 1, we can compare the speed of the electron to the speed of light.
03

Calculating potential difference

To find the potential difference, we can use the formula for kinetic energy of a particle, because this energy was provided by the potential difference that the electron passed through. The kinetic energy of a particle is given by \(K.E. = \frac{1}{2} m v^2\). Thus, the potential difference V can be calculated by \( V = \frac{K.E.}{e} \), where e is the electronic charge, \(1.6 \times 10^{-19} C\). Substituting the calculated speed v from Step 1, and the given values for m and e, we can find the potential difference.

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Key Concepts

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

Double-Slit Interference
Double-slit interference is a fascinating phenomenon that occurs when waves, such as light or in this case, electron waves, pass through two closely spaced slits. The result is an interference pattern of alternating light and dark fringes. This happens because the waves emerging from each slit overlap and combine.
  • Constructive interference occurs when the waves align, creating bright fringes.
  • Destructive interference occurs when the waves are out of sync, resulting in dark fringes.
Electron diffraction experiments, like the one described, use this principle to show that electrons have wave-like properties. The distance between the slits and the observing screen impacts where fringes are seen, as it influences the path difference between the waves from each slit.
In the given exercise, the first dark fringe is noted at \(\pm18.0^{\circ}\), highlighting the destructive interference between the electron waves.
de Broglie Wavelength
The de Broglie wavelength is a concept introduced by physicist Louis de Broglie. It proposes that all matter exhibits wave-like properties, not just light. This wavelength is inversely proportional to the momentum of a particle, given by the formula:\[\lambda = \frac{h}{p}\]where \(h\) is the Planck's constant and \(p\) is the momentum of the particle. For electrons, this concept is crucial to understanding their behavior in quantum mechanics.
  • The exercise assumes a de Broglie wavelength of \(2.43 \times 10^{-12}\ m\) for the electrons, indicating their wave nature.
  • This wavelength plays a part in determining the position of interference fringes in the double-slit experiment.
By using the de Broglie wavelength, we can calculate various aspects of an electron's motion, including its velocity, when related to double-slit interference patterns.
Relativistic Electrons
To understand if electrons are relativistic, we need to evaluate their speed relative to the speed of light. Electrons are considered relativistic when their speed approaches a significant fraction (generally more than 10%) of the speed of light (\(3.00 \times 10^8\ m/s\)). This requires using relativistic physics to accurately describe their behavior.
In the problem, we calculate the speed of the electrons once they have passed through a potential difference and compare that speed to the speed of light.
  • If the calculated speed is a significant percentage of the speed of light, relativistic effects cannot be ignored.
  • Non-relativistic calculations might yield incorrect results at these high speeds.
By checking the calculated velocity against this relativistic threshold, we can determine the necessary physics models to apply when evaluating the behavior of electrons in experiments.

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Most popular questions from this chapter

In the second type of helium-ion microscope, a \(1.2 \mathrm{MeV}\) ion passing through a cell loses \(0.2 \mathrm{MeV}\) per \(\mu \mathrm{m}\) of cell thickness. If the energy of the ion can be measured to \(6 \mathrm{keV},\) what is the smallest difference in thickness that can be discemed? (a) \(0.03 \mu \mathrm{m}\) (b) \(0.06 \mu \mathrm{m}\) (c) \(3 \mu \mathrm{m} ;\) (d) \(6 \mu \mathrm{m}\)

A hydrogen atom is in a state with energy \(-1.51 \mathrm{eV}\). In the Bohr model, what is the angular momentum of the electron in the atom, with respect to an axis at the nucleus?

Imagine another universe in which the value of Planck's constant is \(0.0663 \mathrm{~J} \cdot \mathrm{s},\) but in which the physical laws and all other physical constants are the same as in our universe. In this universe, two physics students are playing catch. They are \(12 \mathrm{~m}\) apart, and one throws a \(0.25 \mathrm{~kg}\) ball directly toward the other with a speed of \(6.0 \mathrm{~m} / \mathrm{s}\). (a) What is the uncertainty in the ball's horizontal momentum, in a direction perpendicular to that in which it is being thrown, if the student throwing the ball knows that it is located within a cube with volume \(125 \mathrm{~cm}^{3}\) at the time she throws it? (b) By what horizontal distance could the ball miss the second student?

A triply ionized beryllium ion, \(\mathrm{Be}^{3+}\) (a beryllium atom with three electrons removed), behaves very much like a hydrogen atom except that the nuclear charge is four times as great. (a) What is the ground-level energy of \(\mathrm{Be}^{3+}\) ? How does this compare to the groundlevel energy of the hydrogen atom? (b) What is the ionization energy of \(\mathrm{Be}^{3+} ?\) How does this compare to the ionization energy of the hydrogen atom? (c) For the hydrogen atom, the wavelength of the photon emitted in the \(n=2\) to \(n=1\) transition is \(122 \mathrm{nm}\) (see Example 39.6 ). What is the wavelength of the photon emitted when a \(\mathrm{Be}^{3+}\) ion undergoes this transition? (d) For a given value of \(n\), how does the radius of an orbit in \(\mathrm{Be}^{3+}\) compare to that for hydrogen?

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