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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

Radiation has been detected from space that is characteristic of an ideal radiator at \(T\) = 2.728 K. (This radiation is a relic of the Big Bang at the beginning of the universe.) For this temperature, at what wavelength does the Planck distribution peak? In what part of the electromagnetic spectrum is this wavelength?

For crystal diffraction experiments (discussed in Section 39.1), wavelengths on the order of 0.20 nm are often appropriate. Find the energy in electron volts for a particle with this wavelength if the particle is (a) a photon; (b) an electron; (c) an alpha particle (\(m\) = 6.64 \(\times\) 10\(^{-27}\) kg).

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?

To investigate the structure of extremely small objects, such as viruses, the wavelength of the probing wave should be about one-tenth the size of the object for sharp images. But as the wavelength gets shorter, the energy of a photon of light gets greater and could damage or destroy the object being studied. One alternative is to use electron matter waves instead of light. Viruses vary considerably in size, but 50 nm is not unusual. Suppose you want to study such a virus, using a wave of wavelength 5.00 nm. (a) If you use light of this wavelength, what would be the energy (in eV) of a single photon? (b) If you use an electron of this wavelength, what would be its kinetic energy (in eV)? Is it now clear why matter waves (such as in the electron microscope) are often preferable to electromagnetic waves for studying microscopic objects?

High-speed electrons are used to probe the interior structure of the atomic nucleus. For such electrons the expression \(\lambda = h/p\) still holds, but we must use the relativistic expression for momentum, \(p = mv/\sqrt{1 - v^2/c^2}\). (a) Show that the speed of an electron that has de Broglie wavelength \(\lambda\) is $$v = \frac{c}{\sqrt1+(mc\lambda/h)^2} $$ (b) The quantity \(h/mc\) equals 2.426 \(\times\) 10\(^{-12}\) m. (As we saw in Section 38.3, this same quantity appears in Eq. (38.7), the expression for Compton scattering of photons by electrons.) If \(\lambda\) is small compared to \(h/mc\), the denominator in the expression found in part (a) is close to unity and the speed \(v\) is very close to c. In this case it is convenient to write \(v = (1 - \Delta)c\) and express the speed of the electron in terms of rather than v. Find an expression for \(\delta\) valid when \(\lambda \ll h mc\). [\(Hint:\) Use the binomial expansion (1 + \(z)^n = 1 + nz + [n(n - 1)z^2/2] + \cdots\) g, valid for the case 0 z 0 6 1.4 (c) How fast must an electron move for its de Broglie wavelength to be 1.00 \(\times\) 10\(^{-15}\) m, comparable to the size of a proton? Express your answer in the form \(v =(1 - \Delta)c\), and state the value of \(\Delta\)

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