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

Expert verified
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

01

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 \(K = \frac{1}{40} \mathrm{eV} = \frac{1}{40} \times 1.6 \times 10^{-19} \mathrm{J}\).
02

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.
03

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.

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