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Chapter 28: Problem 61

The neutrons produced in fission reactors have a wide range of kinetic energies. After the neutrons make several collisions with atoms, they give up their excess kinetic energy and are left with the same average kinetic energy as the atoms, which is \(\frac{3}{2} k_{\mathrm{B}} T .\) If the temperature of the reactor core is \(T=400.0 \mathrm{K},\) find (a) the average kinetic energy of the thermal neutrons, and (b) the de Broglie wavelength of a neutron with this kinetic energy.

Short Answer

Expert verified
Answer: The average kinetic energy of the thermal neutrons is 1.24 × 10^(-20) J, and the de Broglie wavelength is approximately 1.47 × 10^(-10) m.

Step by step solution

01

1. Find the average kinetic energy of the thermal neutrons

We are given the temperature, \(T\), and the formula for the average kinetic energy: \(\text{KE}_{\text{avg}} = \frac{3}{2} k_{\mathrm{B}} T\). We just need to plug in the given temperature and Boltzmann's constant, \(k_{\mathrm{B}} = 1.38 \times 10^{-23} \, \text{J/K}\). \(\text{KE}_{\text{avg}} = \frac{3}{2} (1.38 \times 10^{-23} \, \text{J/K}) (400.0 \mathrm{K})\)
02

2. Calculate the average kinetic energy

Now, we compute the average kinetic energy: \(\text{KE}_{\text{avg}} = 1.24 \times 10^{-20} \, \text{J}\)
03

3. Find the de Broglie wavelength of a neutron

To find the de Broglie wavelength of a neutron, we will use the de Broglie wavelength formula: \(\lambda = \frac{h}{p}\), where \(\lambda\) is the de Broglie wavelength, \(h\) is Planck's constant, \(6.63 \times 10^{-34} \, \text{Js}\), and \(p\) is the momentum of the neutron. To find the kinetic energy, we need the momentum, which in turn depends on the velocity of the neutron. Given the kinetic energy, we can find the neutron's velocity using the formula: \(\text{KE} = \frac{1}{2} m v^2\), where \(m\) is the mass of the neutron, \(\approx 1.67 \times 10^{-27} \, \text{kg}\), and \(v\) is the neutron's velocity. We can solve this equation for the velocity: \(v^2 = \frac{2 \times \text{KE}}{m}\).
04

4. Calculate the neutron's velocity

Now, we can plug in the values and compute the neutron's velocity: \(v^2 = \frac{2 \times 1.24 \times 10^{-20} \, \text{J}} {1.67 \times 10^{-27} \, \text{kg}}\) After solving for \(v\), we get: \(v \approx 2700.8 \, \text{m/s}\)
05

5. Calculate the momentum of the neutron

Now we can compute the momentum of the neutron, as \(p = mv\): \(p = (1.67 \times 10^{-27} \, \text{kg}) (2700.8 \, \text{m/s})\) \(p \approx 4.52 \times 10^{-24} \, \text{kg m/s}\)
06

6. Calculate the de Broglie wavelength

Finally, we can calculate the de Broglie wavelength using the formula \(\lambda = \frac{h}{p}\): \(\lambda = \frac{6.63 \times 10^{-34} \, \text{Js}}{4.52 \times 10^{-24} \, \text{kg m/s}}\) After computing it, we find the de Broglie wavelength to be: \(\lambda \approx 1.47 \times 10^{-10} \, \text{m}\) In summary, the average kinetic energy of the thermal neutrons is \(1.24 \times 10^{-20} \, \text{J}\), and the de Broglie wavelength of a neutron with this kinetic energy is \(\approx 1.47 \times 10^{-10} \, \text{m}\).

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Most popular questions from this chapter

What is the minimum kinetic energy of an electron confined to a region the size of an atomic nucleus \((1.0 \mathrm{fm}) ?\)
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