Video Transcript
In a nuclear fission reactor, the moderator is a material used to slow down the free neutrons in the reactor core, which increases the probability that they will cause a uranium nucleus to undergo fission. The neutrons must have kinetic energies of about 0.0400 electron volts. What is the de Broglie wavelength of a neutron with this kinetic energy? Use a value of 1.67 times 10 to the negative 27 kilograms for the rest mass of a neutron and a value of 6.63 times 10 to the negative 34 joules-seconds for the Planck constant. Give your answer in scientific notation to two decimal places.
Although this question does contain some interesting information about nuclear fission, what we are actually looking for is the de Broglie wavelength of a neutron with a particular kinetic energy. To accomplish this, we are also given the mass of the neutron and a value for the Planck constant. The only other piece of information that we will need is the de Broglie wavelength formula. This formula relates the wavelength and momentum of a particle as wavelength equals the Planck constant divided by momentum.
Now to use this formula, we need the momentum of the particle. But we don’t know the momentum of our neutron. We know its kinetic energy and its mass. Luckily, there is a simple formula that relates these three quantities. For a nonrelativistic particle, the kinetic energy is equal to the square of the momentum divided by two times the mass. And this formula follows directly from kinetic energy equals one-half mass times speed squared. Now, if we use the given energy and mass to find the speed of these neutrons, we would find that their speed is about five orders of magnitude smaller than the speed of light. So the nonrelativistic assumption is valid.
Multiplying both sides by two times the mass, we get that two 𝑚𝐸 equals 𝑃 squared. And taking the square root of both sides, we see that momentum is equal to the square root of two times the mass times the kinetic energy. Substituting this into the de Broglie wavelength formula, we arrive at a formula for the wavelength using only the quantities we are given: the Planck constant, the rest mass of a neutron, and the neutron’s kinetic energy.
Before we substitute values into this expression, note that our mass is given in kilograms and our Planck constant is given in joule-seconds. Both of these are SI base units. So before we plug in values, we should convert the kinetic energy of the neutron given in electron volts to SI base units of joules. One electron volt is 1.60 times 10 to the negative 19 joules. So 0.0400 electron volts is 6.40 times 10 to the negative 21 joules.
Alright, let’s now clear some space so we can substitute values into our expression for the wavelength. We have 6.63 times 10 to the negative 34 joule-seconds divided by the square root of two times 1.67 times 10 to the negative 27 kilograms times 6.40 times 10 to the negative 21 joules. When we plug the numbers into a calculator and round to two decimal places, we get 1.43 times 10 to the negative 10. And the units of this answer are meters, which we know because we’re calculating a wavelength which has units of length. And all of the units that we have used, joules, seconds, and kilograms, are SI base units. So any combination of them must also produce an SI base unit, in this case meters, the SI base unit for length. So the neutrons in this reactor have de Broglie wavelengths of 1.43 times 10 to the negative 10 meters.
