Question

The neutrons in a parallel beam, each having kinetic energy \(1 / 40 \mathrm{eV}\) (which is approximately corresponding to “room temperature"), are directed through two slits \(0.50 \mathrm{~mm}\) apart. How far apart will the interference peaks be on a screen \(1.5 \mathrm{~m}\) away?

Short answer

Answer: To find the distance between the interference peaks on the screen, we need to follow these steps: 1. Calculate the de Broglie wavelength of the neutrons using their kinetic energy and the formulas for kinetic energy and the de Broglie wavelength. 2. Use the double-slit interference formula, the small angle approximation, and the calculated de Broglie wavelength to find the distance between the interference peaks on the screen. After following these steps and substituting the given values into the formulas, we can find the distance between the interference peaks on the screen.

Step-by-step solution

Find the de Broglie wavelength of the neutrons

Given the kinetic energy of the neutrons, we can find their momentum and subsequently their de Broglie wavelength using the following formula: $$ \lambda = \frac{h}{p} $$ Where \(\lambda\) is the de Broglie wavelength, \(h\) is the Planck's constant (\(6.626 \times 10^{-34} \mathrm{Js}\)), and \(p\) is the momentum of the neutron. To find the momentum, we need the mass and velocity of the neutron. We can find the velocity using the kinetic energy formula: $$ K = \frac{1}{2}mv^2 $$ Where \(K\) is the kinetic energy, \(m\) is the mass, and \(v\) is the velocity. For a neutron, the mass is approximately \(m = 1.675 \times 10^{-27} \mathrm{kg}\), and the given kinetic energy is \(K = \frac{1}{40} \mathrm{eV} = \frac{1}{40} \times 1.6 \times 10^{-19} \mathrm{J}\).

Use the double-slit interference formula to find the distance between the interference peaks

The formula for double-slit interference is given by: $$ d \sin{\theta} = m\lambda $$ Where \(d\) is the separation between the two slits, \(m\) is an integer representing the peak order, and \(\theta\) is the angle of the interference peak from the central maximum. Given the small angle approximation, we can use the following relationship: $$ \tan{\theta} \approx \sin{\theta} \approx \theta \approx \frac{y}{L} $$ Where \(y\) is the distance between the interference peaks on the screen, and \(L\) is the distance between the slits and the screen. In this problem, \(d = 0.50 \mathrm{mm} = 0.50 \times 10^{-3} \mathrm{m}\), and \(L=1.5 \mathrm{m}\). We can now find \(y\) using the formula \(y = L \tan{\theta} = L \frac{\lambda}{d}\), where \(\lambda\) is the de Broglie wavelength we calculated in Step 1.

Calculate the distance between interference peaks

Now that we have all the variables, we can plug them into the formula: $$ y = L \frac{\lambda}{d} $$ After substituting the values for \(L\), \(\lambda\), and \(d\), we will find the distance \(y\) between the interference peaks on the screen.

Key concepts

double-slit interference

Double-slit interference is a fascinating phenomenon that arises when waves pass through two narrow slits and then overlap on the other side, creating a pattern of bright and dark spots, known as interference patterns. This was first famously demonstrated by Thomas Young in the early 1800s with light waves. However, this principle is not limited to light. It applies to any wave, including sound, water, and even particles such as neutrons, as described in this exercise.
In the specific context of this exercise, neutrons, despite being particles, exhibit wave-like behavior. When a beam of neutrons passes through two closely spaced slits, the individual quantum waves associated with each neutron can interfere with one another. This results in a pattern of alternating bright (constructive interference) and dark (destructive interference) regions over a screen placed on the other side.
The formula used to determine the positions of these interference peaks is given by:

• \[d \sin{\theta} = m\lambda\]

This helps us locate where the peaks of the interference pattern will appear, depending on the slit's separation \(d\), the order \(m\), and the wavelength \(\lambda\) of the neutrons. By applying the small angle approximation, a convenient simplification, we express \(\theta\) in terms of \(y\) and \(L\):

• \[\theta \approx \frac{y}{L}\]

This equation makes it easy to calculate distances on the screen based on the de Broglie wavelength and slit geometry.

neutron physics

Neutrons are subatomic particles, famously known for their neutral charge, which makes them different from protons and electrons. In physics, neutrons are incredibly important for nuclear reactions and processes, such as fission. When dealing with neutron beams, we often consider their behavior in terms of their wave and particle duality.
This exercise explores this duality by considering the wave-like properties of neutrons. Each neutron, despite being a particle, has an associated wave known as a matter wave or de Broglie wave. This is fundamental to quantum mechanics and supports theories where every particle has a dual-wave nature.
The de Broglie wavelength, \(\lambda\), of a neutron is given by the formula:

• \[\lambda = \frac{h}{p}\]

Here, \(h\) is Planck's constant (6.626 \times 10^{-34} \mathrm{Js}), and \(p\) is the momentum of the neutron. Momentum \(p\), for neutrons, can be determined from their kinetic properties, as is done in the given exercise. It's through this wavelength that neutrons demonstrate wave-like interference patterns when interacting with obstacles, such as the double-slit setup described.
Neutrons, being neutrally charged, are valuable in experiments involving interference, as they are less likely to interact electromagnetically with particles than charged particles. This property helps in making precise measurements in experimental physics.

kinetic energy

Kinetic energy is the energy possessed by any object by virtue of its motion. It's a critical concept in physics, measured in Joules in SI units, and can be calculated using the well-known formula:

• \[K = \frac{1}{2}mv^2\]

where \(m\) is the mass and \(v\) is the velocity of the object. This exercise considers neutrons, where the kinetic energy is given in electronvolts (eV), a common unit in atomic and particle physics.
A neutron's kinetic energy allows us to determine its velocity, which is necessary to find its momentum, and subsequently its de Broglie wavelength. In this exercise, the kinetic energy per neutron is given as \(1/40 \mathrm{eV}\). To link this to standard units, remember that \(1 \, ext{eV} = 1.6 \, \times 10^{-19} \, ext{Joule}\).
Since the neutrons are moving, they possess this kinetic energy, which contributes to their ability to interfere with each other upon passing through the slits. It is this energy that makes it possible to visualize the wave properties of particles on a macroscopic scale when performing the double-slit experiment.
The kinetic energy of particles, such as neutrons, is a key aspect in providing insights into both classical and quantum mechanics arenas. It helps bridge the understanding of wave-particle duality, a cornerstone of modern physics.