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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?

Short Answer

Expert verified
Relativistic formulas are not needed, and the slit width is approximately 1.25 nm.

Step by step solution

01

Determine Electron Speed

To determine if we need to use relativity formulas, we first find the speed of the electrons. The kinetic energy gained by the electrons is equal to the potential energy given by the potential difference:\[ KE = eV = \frac{1}{2}mv^2 \]Where \(e\) is the electron charge (\(1.6 \times 10^{-19} \, \text{C}\)), \(V\) is the potential difference (0.100 kV = 100 V), \(m\) is the mass of the electron (\(9.11 \times 10^{-31} \, \text{kg}\)), and \(v\) is the velocity of the electron. Solve for \(v\):\[ v = \sqrt{\frac{2eV}{m}} \]Substitute the known values to find \(v\).
02

Evaluate Need for Relativity

With the velocity from Step 1, determine the need for relativistic equations by comparing the electron speed to a significant fraction of the speed of light (approximately 0.1c). If \(v\) is less than 0.1c, classical mechanics is sufficient. Otherwise, use relativistic formulas. Calculate \(0.1c\) where the speed of light \(c = 3 \times 10^8 \, \text{m/s}\).
03

Apply Diffraction Formula

Assuming classical physics is sufficient, use the diffraction formula for the first minimum:\[ a \sin \theta = n\lambda \]where \(a\) is the slit width, \(\theta = 14.6^\circ\), \(n = 1\) (first minimum), and \(\lambda\) is the de Broglie wavelength of electrons:\[ \lambda = \frac{h}{mv} \]with Planck's constant \(h = 6.626 \times 10^{-34} \, \text{Js}\). Use \(v\) from Step 1 to calculate \(\lambda\), then solve for \(a\).
04

Calculate Slit Width

Using the values of \(\sin \theta\) and \(\lambda\) derived in earlier steps, substitute into the diffraction formula to find \(a\):\[ a = \frac{n\lambda}{\sin \theta} \]Calculate \(\sin 14.6^\circ\) and substitute. Calculate \(a\), the slit width, in meters or centimeters as desired.

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Key Concepts

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

Kinetic Energy
In this exercise, understanding the concept of kinetic energy is crucial to analyze the movement of electrons. When electrons are accelerated through a potential difference, they gain kinetic energy. This energy can be expressed as a function of the charge of the electron and the potential difference applied, represented by the equation:\[ KE = eV \]where:
  • \( KE \) is the kinetic energy
  • \( e \) is the elementary charge, approximately \( 1.6 \times 10^{-19} \, \text{C} \)
  • \( V \) is the potential difference
The kinetic energy gained allows us to find the velocity of the electrons using the expression:\[ KE = \frac{1}{2}mv^2 \]Rearranging gives the velocity:\[ v = \sqrt{\frac{2eV}{m}} \]Here, \( m \) is the mass of an electron, approximately \( 9.11 \times 10^{-31} \, \text{kg} \). Calculating kinetic energy is fundamental for determining the electron speed and examining if relativistic effects are significant.
de Broglie Wavelength
The de Broglie wavelength is a key concept in quantum mechanics linking momentum to wave-like properties of particles. According to de Broglie's hypothesis, particles such as electrons exhibit wave-like characteristics, described by their wavelength:\[ \lambda = \frac{h}{mv} \]where:
  • \( \lambda \) is the de Broglie wavelength
  • \( h \) is Planck's constant, \( 6.626 \times 10^{-34} \, \text{Js} \)
  • \( m \) is the mass of the electron
  • \( v \) is the speed of the electron obtained previously
Understanding the de Broglie wavelength is essential in calculating diffraction patterns, like in this exercise where electrons pass through a slit. Their wavelengths contribute to the interference patterns observed, much like light waves, but on a quantum scale. The computed wavelength helps determine the positions of diffraction minima when electrons interact with the slit.
Relativistic Mechanics
Assessing whether to use relativistic mechanics is crucial in electron beam problems. Electrons, when accelerated to high speeds, may need relativistic equations if their velocity nears a significant fraction of light speed. This scenario occurs when an electron's speed is a considerable portion of the speed of light, approximately:\[ 0.1c \]Here, \( c \) is the speed of light, \( 3 \times 10^8 \, \text{m/s} \). In this specific exercise, if the calculated velocity surpasses this threshold, relativistic adjustments are necessary. This would involve using the relativistic kinetic energy formula:\[ KE = (\gamma - 1)mc^2 \]with \( \gamma \) being the Lorentz factor:\[ \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \]However, if the electron speed is much less than \( 0.1c \), classical mechanics provides sufficient accuracy for all computations.
Diffraction Minimum
The concept of diffraction minimum is central to understanding how electrons create a diffraction pattern after passing through a slit. When waves, including particle waves like electrons, pass through slits, they exhibit behavior characterized by constructive and destructive interference. For a diffraction minimum, where no or minimal wave intensity is observed, the condition is given by the formula:\[ a \sin \theta = n\lambda \]where:
  • \( a \) is the slit width
  • \( \theta \) is the angle at which minimum occurs
  • \( n \) is the order of the minimum, with 1 being the first minimum
  • \( \lambda \) is the wavelength of electrons as per de Broglie's relation
To find the slit width, the wavelength calculated before is used along with the observed angle \( \theta = 14.6^\circ \), leading to solving for \( a \). Knowing diffraction minima aids in measuring slit dimensions and understanding wave-particle duality in quantum mechanics.

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Most popular questions from this chapter

Why is it easier to use helium ions rather than neutral helium atoms in such a microscope? (a) Helium atoms are not electrically charged, and only electrically charged particles have wave properties. (b) Helium atoms form molecules, which are too large to have wave properties. (c) Neutral helium atoms are more difficult to focus with electric and magnetic fields. (d) Helium atoms have much larger mass than helium ions do and thus are more difficult to accelerate.

(a) A nonrelativistic free particle with mass \(m\) has kinetic energy \(K\). Derive an expression for the de Broglie wavelength of the particle in terms of \(m\) and \(K\). (b) What is the de Broglie wavelength of an 800-eV electron?

An atom with mass \(m\) emits a photon of wavelength \(\lambda\). (a) What is the recoil speed of the atom? (b) What is the kinetic energy \(K\) of the recoiling atom? (c) Find the ratio \(K/E\), where \(E\) is the energy of the emitted photon. If this ratio is much less than unity, the recoil of the atom can be neglected in the emission process. Is the recoil of the atom more important for small or large atomic masses? For long or short wavelengths? (d) Calculate \(K\) (in electron volts) and \(K/E\) for a hydrogen atom (mass 1.67 \(\times\) 10\(^{-27}\)kg) that emits an ultraviolet photon of energy 10.2 eV. Is recoil an important consideration in this emission process?

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?

The wavelength 10.0 \(\mu\)m is in the infrared region of the electromagnetic spectrum, whereas 600 nm is in the visible region and 100 nm is in the ultraviolet. What is the temperature of an ideal blackbody for which the peak wavelength \(\lambda_m\) is equal to each of these wavelengths?

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