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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?

Short Answer

Expert verified
(a) Electrons are not relativistic, as energy is much less than rest energy. (b) Potential difference is approximately 54.0 V.

Step by step solution

01

Determine the Wavelength of Electrons Using Interference Formula

For the dark fringe in a double-slit experiment, the condition is given by \( d \sin \theta = (m + \frac{1}{2}) \lambda \), where \( d = 1.25\, \text{nm} \) is the slit separation, \( m \) is the order of the dark fringe, \( \theta = 18.0^\circ \), and \( \lambda \) is the wavelength. For the first dark fringe, \( m = 0 \). Therefore, \( \lambda = \frac{1.25 \times 10^{-9} \sin 18.0}{0.5} \approx 7.69 \times 10^{-10}\, \text{m} \).
02

Check if Electrons are Relativistic

Use de Broglie's equation \( \lambda = \frac{h}{p} \) to find the momentum \( p \). Assuming \( \lambda = 7.69 \times 10^{-10}\, \text{m} \), and \( h = 6.626 \times 10^{-34}\, \text{J}\cdot\text{s} \), \( p = \frac{6.626 \times 10^{-34}}{7.69 \times 10^{-10}} \approx 8.62 \times 10^{-25} \text{kg}\cdot\text{m/s} \). Assuming non-relativistic conditions: \( p = mv \) and \( v = \sqrt{\frac{2qV}{m}} \), compare the speed, \( v \), to the speed of light, \( c \), to establish if relativistic effects are significant. As \( v \) approaches \( c \), electrons are relativistic.
03

Calculate the Potential Difference Using Energy Concept

Under non-relativistic approximation, kinetic energy \( KE = \frac{1}{2}mv^2 = qV \). Rearranging gives \( V = \frac{p^2}{2mq} \), using \( m = 9.11 \times 10^{-31} \text{kg} \) and \( q = 1.602 \times 10^{-19} \text{C} \), \( V = \frac{{(8.62 \times 10^{-25})}^2}{2 \times 9.11 \times 10^{-31} \times 1.602 \times 10^{-19}} \approx 54.0 \text{V} \).
04

Analyze Relativistic Nature of Electrons Again

Since potential difference \( V = 54.0 \text{V} \) leads to electron energy \( eV = (54 \text{V})(1.602 \times 10^{-19} \text{C}) = 8.66 \times 10^{-18} \text{J} \) which is far less than rest energy \( mc^2 = 8.19 \times 10^{-14} \text{J} \), thus relativistic effects are negligible.

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Key Concepts

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

de Broglie's equation
De Broglie's equation is a fundamental concept that connects the wave and particle nature of matter. It posits that every moving particle or object has an associated wave. This wave is sometimes called a "matter wave." The equation is expressed as: \( \lambda = \frac{h}{p} \), where \( \lambda \) is the wavelength, \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \)), and \( p \) is the momentum of the particle.

In our exercise, we determined that the electrons have a wavelength of approximately \( 7.69 \times 10^{-10} \, \text{m} \). This calculation involved the measurement of interference patterns, illustrating how de Broglie's hypothesis manifests in real-world experiments. By calculating the momentum of the electrons using the wavelength formula, we can understand their behavior in terms of both waves and particles.

The concept is crucial in quantum mechanics because it helps explain phenomena that classical theories couldn't, such as electron interference patterns in a double-slit experiment.
double-slit experiment
The double-slit experiment is a classic demonstration of the wave nature of light and particles. When a wavefront encounters two slits close together, it creates interference patterns on a screen beyond the slits. It provides evidence of wave-particle duality, a central concept of quantum mechanics.

In this particular exercise, electrons passed through the slits, producing a pattern of dark and light fringes on a screen. The dark fringes occur due to destructive interference, where wave crests meet wave troughs and cancel each other out. The formula \( d \sin \theta = (m + \frac{1}{2}) \lambda \) describes the location of these interference fringes. Here, \( d \) is the distance between slits, \( \theta \) is the angle at which the fringe occurs, \( m \) is the fringe order, and \( \lambda \) is the wavelength.

By calculating the angle for the first dark fringe, we derive important information about the wave properties of electrons, reinforcing the principles explored by de Broglie.
potential difference
The concept of potential difference is essential in describing how kinetic energy is imparted to charged particles. When an electric potential difference, or voltage, is applied, charges gain energy. For electrons, the energy gained is often expressed as \( eV \), where \( e \) is the electron charge and \( V \) is the potential difference.

In the context of accelerating electrons through a potential difference, they start from rest and gain kinetic energy as they travel through the electric field. This energy helps calculate their speed and their potential to exhibit relativistic behavior.

In our exercise, the potential difference was calculated to be approximately 54.0 Volts, which was insufficient to make the electrons relativistic. This conclusion is based on comparing their kinetic energy to their rest energy, \( mc^2 \). Therefore, at such a potential, relativistic effects are negligible, simplifying the calculations and allowing classical approaches to be sufficient.

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Most popular questions from this chapter

An electron is moving with a speed of 8.00 \(\times\) 10\(^6\) m/s. What is the speed of a proton that has the same de Broglie wavelength as this electron?

(a) An atom initially in an energy level with \(E\) = -6.52 eV absorbs a photon that has wavelength 860 nm. What is the internal energy of the atom after it absorbs the photon? (b) An atom initially in an energy level with \(E\) = -2.68 eV emits a photon that has wavelength 420 nm. What is the internal energy of the atom after it emits the photon?

A pesky 1.5-mg mosquito is annoying you as you attempt to study physics in your room, which is 5.0 m wide and 2.5 m high. You decide to swat the bothersome insect as it flies toward you, but you need to estimate its speed to make a successful hit. (a) What is the maximum uncertainty in the horizontal position of the mosquito? (b) What limit does the Heisenberg uncertainty principle place on your ability to know the horizontal velocity of this mosquito? Is this limitation a serious impediment to your attempt to swat it?

Imagine another universe in which the value of Planck's constant is 0.0663 J \(\cdot\) 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 m apart, and one throws a 0.25-kg ball directly toward the other with a speed of 6.0 m/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 cm\(^3\) at the time she throws it? (b) By what horizontal distance could the ball miss the second student?

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