Video Transcript
An electron has a rest mass of 9.11 times 10 to the negative 31 kilograms. If the electron has a kinetic energy of 1.14 times 10 to the negative 27 joules, what is its de Broglie wavelength? Use a value of 6.63 times 10 to the negative 34 joule-seconds for the Planck constant. Give your answer in scientific notation to two decimal places.
In this question, we are looking for a de Broglie wavelength. So the first thing that comes to mind is the de Broglie relation that relates a particle’s wavelength to its momentum. This relation is that 𝜆, the wavelength, is equal to ℎ, the Planck constant, divided by 𝑝, the particle’s momentum. Now, we know the Planck constant, and we are looking for the wavelength. But we don’t know the momentum of our electron. All we know are its rest mass, 9.11 times 10 to the negative 31 kilograms, and its kinetic energy, 1.14 times 10 to the negative 27 joules.
So we need to calculate momentum from rest mass and kinetic energy. Since the electron’s kinetic energy is so small, we can assume that it is nonrelativistic and write kinetic energy equals one-half mass times speed squared. In fact, if we used this formula to calculate the speed, we would find that the electron is moving at only about 50 meters per second, much, much slower than the speed of light. And so, our nonrelativistic assumption is justified. But we don’t need the speed of the electron. We need its momentum, which for nonrelativistic speeds can be written as 𝑝 equals 𝑚𝑣: momentum equals mass times speed.
Looking back at our formula for kinetic energy, we see that if we had an extra factor of mass on the right-hand side, we would have 𝑚 squared times 𝑣 squared, which is 𝑚𝑣 squared, which is 𝑝 squared. To get this factor of 𝑚, all we have to do is multiply both sides of this equation by mass. So mass times kinetic energy is equal to one-half 𝑚 squared 𝑣 squared, which is the same thing as one-half 𝑚𝑣 squared, which is one-half times the square of the momentum of the particle. Now, multiply both sides by two to find that two times the mass times the kinetic energy of the particle is equal to the square of the particle’s momentum.
Now take the square root of both sides to isolate 𝑝. 𝑝 is equal to the square root of two times 𝑚 times the kinetic energy. This gives us momentum in terms of rest mass and kinetic energy, which are the two quantities we are given in the question. Substituting this formula into the de Broglie relation, we have 𝜆 equals ℎ divided by the square root of two times 𝑚 times KE, which gives us wavelength in terms of quantities that we are given in the question. Alright, all that’s left to do is substitute 9.11 times 10 to the negative 31 kilograms for 𝑚, 1.14 times 10 to the negative 27 joules for KE, and 6.63 times 10 to the negative 34 joule-seconds for ℎ.
Let’s clear some space to do this calculation. When we plug all of these numbers into a calculator, we come up with 1.4547 and several more decimal places times 10 to the negative five. As for the units, we could directly compute joules times seconds divided by the square of kilograms times joules. But we could also recognize that our final result must have units of length because our final result is a wavelength. And joules and seconds and kilograms are all SI base units. So the result will be the SI base unit for length, which is meters.
We now have a numerical answer expressed in scientific notation with appropriate units. All that’s left is to round to two decimal places. 1.454 etcetera, rounded to only two decimal places, is 1.45. So the de Broglie wavelength of our electron is 1.45 times 10 to the negative five meters.
