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Coherent light is passed through two narrow slits whose separation is 20.0 \(\mu\)m. The second-order bright fringe in the interference pattern is located at an angle of 0.0300 rad. If electrons are used instead of light, what must the kinetic energy (in electron volts) of the electrons be if they are to produce an interference pattern for which the second-order maximum is also at 0.0300 rad?

Short Answer

Expert verified
The electrons must have a kinetic energy of approximately 9,432 eV.

Step by step solution

01

Understanding the Double Slit Experiment

In a double-slit experiment, coherent waves such as light or electron waves create interference patterns through constructive interference, leading to bright fringes. The positions of these fringes depend on the wavelength of the incident light or particles and the separation between slits.
02

Using the Interference Formula for Bright Fringes

The formula for the angle for bright (constructive interference) fringes is given by: \[ d \sin(\theta) = m\lambda \]where \(d\) is the distance between the slits, \( \theta \) is the angle with the central maximum, \( m \) is the order of the maximum (integer), and \( \lambda \) is the wavelength.
03

Calculating the Wavelength for the Second-Order Maximum

Given \( m = 2 \) (second order), \( \theta = 0.0300 \) rad, and \( d = 20.0 \mu m = 20.0 \times 10^{-6} \) m:Substitute into the interference formula:\[ 20.0 \times 10^{-6} \sin(0.0300) = 2\lambda \]Solving for \( \lambda \), we find \( \lambda = 3.0 \times 10^{-8} \) m.
04

Relating Electron Wavelength to Kinetic Energy

The de Broglie wavelength formula relates the wavelength \( \lambda \) of a particle to its momentum \( p \):\[ \lambda = \frac{h}{p} \]where \( h = 6.626 \times 10^{-34} \) Js is Planck's constant. For electrons, momentum can be expressed as \( p = \sqrt{2m_e K} \), with \( m_e = 9.109 \times 10^{-31} \) kg (electron mass) and \( K \) as kinetic energy.
05

Solving for Electron Kinetic Energy

Substitute for momentum in the de Broglie formula:\[ \lambda = \frac{h}{\sqrt{2m_e K}} \]Rearrange to solve for \( K \):\[ K = \frac{h^2}{2m_e \lambda^2} \]Substitute \( \lambda = 3.0 \times 10^{-8} \) m and known constants:\[ K = \frac{(6.626 \times 10^{-34})^2}{2 \times 9.109 \times 10^{-31} \times (3.0 \times 10^{-8})^2} \]Calculating gives \( K = 1.51 \times 10^{-3} \) Joules, which converts to electron volts (1 eV = 1.602 \times 10^{-19} J):\[ K = \frac{1.51 \times 10^{-3}}{1.602 \times 10^{-19}} \approx 9,432 \text{ eV} \]
06

Final Answer

The kinetic energy of the electrons needed to produce the same second-order maximum at an angle of 0.0300 rad is approximately \(9,432\) eV.

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Key Concepts

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

Interference Patterns
The concept of interference patterns is central to understanding the double slit experiment. These patterns occur when waves overlap, creating regions of constructive and destructive interference. Constructive interference happens when the peaks of waves coincide, resulting in amplified brightness, while destructive interference occurs when peaks and troughs collide, canceling each other out. This phenomenon was historically observed with light, leading to a series of bright and dark regions called fringes.
Imagine a pond where two stones are dropped at once. The overlapping ripples create a similar pattern to that of the interference pattern seen through the slits. This is a hallmark of wave behavior and is evidence of wave-like properties in particles such as electrons.
de Broglie Wavelength
The de Broglie wavelength bridges the wave-particle duality of matter, proposing that particles have wave properties. This is given by the formula: \[ \lambda = \frac{h}{p} \] where \( \lambda \) is the wavelength, \( h \) is Planck's constant, and \( p \) is the momentum of the particle. This concept was groundbreaking in physics because it introduced the idea that not just light, traditionally viewed as a wave, but also particles such as electrons have wavelengths. Understanding an electron's de Broglie wavelength helps explain phenomena like diffraction and interference when they pass through the slits.
Electron Kinetic Energy
To determine the kinetic energy of electrons in experiments like those involving interference patterns, we use the relationship between kinetic energy and momentum given by the de Broglie and classical physics, which is integrated in the formula: \[ K = \frac{h^2}{2m_e \lambda^2} \] where \( K \) is the kinetic energy, \( m_e \) is the mass of the electron, and \( \lambda \) is the de Broglie wavelength. Calculating this gives the energy in joules but we often express it in electron volts (eV) for convenience in quantum experiments. In this scenario, knowing the wavelength helps us find the specific energy electrons need to create a given interference pattern, providing a link between theory and experimental evidence.
Constructive Interference
Constructive interference occurs when the crests of two waves align closely, making the resultant wave amplitude larger. In the context of the double slit experiment, this is seen as bright fringes on the interference pattern.
The formula for the condition of constructive interference is: \[ d \sin(\theta) = m\lambda \] where \( d \) is the distance between slits, \( \theta \) is the angle of observation, \( m \) is the order of the maximum (whole number), and \( \lambda \) is the wavelength. Recognizing the conditions for constructive interference helps predict where the bright spots form and aligns with understanding the underlying wave mechanics of particles. This ability to predict and confirm the existence of interference patterns through constructive interference is fundamentally significant in verifying wave-particle duality.

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Most popular questions from this chapter

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 \(\Delta\)

The silicon-silicon single bond that forms the basis of the mythical silicon- based creature the Horta has a bond strength of 3.80 eV. What wavelength of photon would you need in a (mythical) phasor disintegration gun to destroy the Horta?

(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?

How many photons per second are emitted by a 7.50-mW CO\(_2\) laser that has a wavelength of 10.6 \(\mu\)m?

(a) What accelerating potential is needed to produce electrons of wavelength 5.00 nm? (b) What would be the energy of photons having the same wavelength as these electrons? (c) What would be the wavelength of photons having the same energy as the electrons in part (a)?

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