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Chapter 6: Problem 151

Experiments are performed with ultracold neutrons having velocities as small as \(1.00 \mathrm{m} / \mathrm{s}\). Find the wavelength of such an ultracold neutron and its kinetic energy.

Short Answer

Expert verified
The wavelength of the ultracold neutron is approximately \(3.97 \times 10^{-10} \mathrm{m}\), and its kinetic energy is approximately \(8.34 \times 10^{-28} \mathrm{J}\).

Step by step solution

01

Recall the de Broglie wavelength formula

The de Broglie wavelength of a particle is given by the formula: \(\lambda = \frac{h}{p}\) where \(\lambda\) is the wavelength, \(h\) is the Planck's constant, and \(p\) is the momentum of the particle.
02

Calculate the momentum of the neutron

The momentum of the neutron can be calculated using the formula: \(p = mv\) where \(m\) is the mass of the neutron and \(v\) is its velocity. We are given that the velocity \(v=1.00 \mathrm{m/s}\), and the mass of a neutron is approximately \(m = 1.67 \times 10^{-27} \mathrm{kg}\). Thus, the momentum of the neutron is: \(p = (1.67 \times 10^{-27} \mathrm{kg})(1.00 \mathrm{m/s}) = 1.67 \times 10^{-27} \mathrm{kg \cdot m/s}\)
03

Calculate the wavelength of the neutron

Now, we can use the de Broglie wavelength formula and the calculated momentum to find the wavelength of the neutron. The Planck's constant is approximately \(h = 6.63 \times 10^{-34} \mathrm{J \cdot s}\). Thus, the wavelength is: \(\lambda = \frac{6.63 \times 10^{-34} \mathrm{J \cdot s}}{1.67 \times 10^{-27} \mathrm{kg \cdot m/s}} \approx 3.97 \times 10^{-10} \mathrm{m}\)
04

Calculate the kinetic energy of the neutron

To find the kinetic energy of the neutron, we can use the classical kinetic energy formula: \(K = \frac{1}{2}mv^2\) Using the mass and velocity of the neutron given earlier, we can calculate the kinetic energy: \(K = \frac{1}{2}(1.67 \times 10^{-27} \mathrm{kg})(1.00 \mathrm{m/s})^2 \approx 8.34 \times 10^{-28} \mathrm{J}\)
05

State the results

The wavelength of the ultracold neutron is approximately \(3.97 \times 10^{-10} \mathrm{m}\), and its kinetic energy is approximately \(8.34 \times 10^{-28} \mathrm{J}\).

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Most popular questions from this chapter

What is the de Broglie wavelength of a 10-kg football player running at a speed of \(8.0 \mathrm{m} / \mathrm{s}\) ?

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