Math Problem Statement

The neutrons in a parallel beam, each having kinetic energy 0.026 eV , are directed through two slits 0.40 mm apart.

Part A 

How far apart will the interference peaks be on a screen 1.3 m away? [Hint: First find the wavelength of the neutron.]

Express your answer to two significant figures and include the appropriate units.

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Δx =

Solution

To solve this problem, we need to determine the spacing of the interference peaks Δx\Delta x using the de Broglie wavelength of the neutron. Here's the step-by-step solution:


Step 1: Calculate the de Broglie wavelength of the neutron

The de Broglie wavelength λ\lambda is given by:

λ=hp\lambda = \frac{h}{p}

where:

  • h=6.626×1034J\cdotpsh = 6.626 \times 10^{-34} \, \text{J·s} (Planck's constant)
  • p=2mKp = \sqrt{2 m K} (momentum of the neutron)

Here, mm is the mass of the neutron (m=1.675×1027kgm = 1.675 \times 10^{-27} \, \text{kg}) and KK is the kinetic energy of the neutron (K=0.026eVK = 0.026 \, \text{eV}). Convert KK to joules:

K=0.026eV×1.602×1019JeV=4.1652×1021JK = 0.026 \, \text{eV} \times 1.602 \times 10^{-19} \, \frac{\text{J}}{\text{eV}} = 4.1652 \times 10^{-21} \, \text{J}

Now, calculate the momentum pp:

p=2mK=2(1.675×1027kg)(4.1652×1021J)p = \sqrt{2 m K} = \sqrt{2 (1.675 \times 10^{-27} \, \text{kg})(4.1652 \times 10^{-21} \, \text{J})}

p3.736×1024kg\cdotpm/sp \approx 3.736 \times 10^{-24} \, \text{kg·m/s}

Finally, find λ\lambda:

λ=hp=6.626×10343.736×1024\lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34}}{3.736 \times 10^{-24}}

λ1.774×1010m(or 0.1774nm)\lambda \approx 1.774 \times 10^{-10} \, \text{m} \, (\text{or } 0.1774 \, \text{nm})


Step 2: Calculate the spacing of the interference peaks Δx\Delta x

The formula for the spacing of the interference peaks is:

Δx=λLd\Delta x = \frac{\lambda L}{d}

where:

  • λ=1.774×1010m\lambda = 1.774 \times 10^{-10} \, \text{m} (calculated above)
  • L=1.3mL = 1.3 \, \text{m} (distance to the screen)
  • d=0.40mm=0.40×103md = 0.40 \, \text{mm} = 0.40 \times 10^{-3} \, \text{m}

Substitute the values:

Δx=(1.774×1010)(1.3)0.40×103\Delta x = \frac{(1.774 \times 10^{-10})(1.3)}{0.40 \times 10^{-3}}

Δx5.77×104m(or 0.577mm)\Delta x \approx 5.77 \times 10^{-4} \, \text{m} \, (\text{or } 0.577 \, \text{mm})


Final Answer:

Δx=0.58mm\Delta x = 0.58 \, \text{mm}


Let me know if you would like more details or further clarifications!


Related Questions:

  1. How is the de Broglie wavelength related to the kinetic energy of a particle?
  2. What happens to the interference pattern if the distance between the slits is reduced?
  3. How does the wavelength of a neutron compare to that of a photon of the same energy?
  4. Why is the neutron beam considered as a wave in this context?
  5. What are some practical applications of neutron interference experiments?

Tip: Always ensure energy units are consistent when using physical formulas. Convert eV to joules if necessary!

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Math Problem Analysis

Mathematical Concepts

Wave-Particle Duality
Interference Patterns
de Broglie Wavelength
Kinetic Energy and Momentum

Formulas

de Broglie wavelength: λ = h / p
Momentum: p = √(2mK)
Interference spacing: Δx = (λL) / d

Theorems

de Broglie's hypothesis (relating wavelength to momentum)
Young's Double-Slit Experiment (interference patterns)

Suitable Grade Level

Undergraduate Physics or Advanced High School (Grades 11-12)