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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

A beam of electrons is accelerated from rest and then passes through a pair of identical thin slits that are 1.25 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?

(a) What is the energy of a photon that has wavelength 0.10 \(\mu\)m ? (b) Through approximately what potential difference must electrons be accelerated so that they will exhibit wave nature in passing through a pinhole 0.10 \(\mu\)m in diameter? What is the speed of these electrons? (c) If protons rather than electrons were used, through what potential difference would protons have to be accelerated so they would exhibit wave nature in passing through this pinhole? What would be the speed of these protons?

An electron has a de Broglie wavelength of 2.80 \(\times\) 10\(^{-10}\) m. Determine (a) the magnitude of its momentum and (b) its kinetic energy (in joules and in electron volts).

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

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.

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