### Video Transcript

What is the de Broglie wavelength of a proton whose kinetic energy is 2.0 megaelectron volts?

Let’s call this kinetic energy of 2.0 megaelectron volts capital KE. We want to solve for the de Broglie wavelength of this proton; we’ll call that 𝜆. To start, let’s recall the mathematical relationship for the de Broglie wavelength of a particle. A particle’s de Broglie wavelength is equal to ℎ, which is a constant called Planck’s constant, which has a value of approximately 6.626 times 10 to the negative 34th square meters kilograms per second.

And to solve for the de Broglie wavelength, we divide this constant by 𝑝, the momentum of our particle. When we recall that momentum is equal to an object’s mass times its velocity, 𝑣, that means we can apply the de Broglie wavelength equation to our scenario by writing that the wavelength of the proton equals Planck’s constant divided by the proton’s mass times its speed.

However, we weren’t given the proton’s speed but we were given its kinetic energy. And in general, the kinetic energy of an object, KE, is equal to one half the mass of the object times its speed squared. This means that if we were to solve for an object’s speed in terms of its kinetic energy, then that speed is equal to the square root of two times its overall kinetic energy divided by 𝑚.

So we can rewrite our equation for the de Broglie wavelength of a proton by substituting in for 𝑣. The factors of the proton mass in the denominator simplify and we can write that the de Broglie wavelength from the proton is Planck’s constant divided by the square root of two times the proton’s mass times its kinetic energy.

Looking up the mass of a proton in a table, we find that it’s approximately equal to 1.67 times 10 to the negative 27th kilograms. Now that we have values for ℎ, 𝑚, and KE, we’re ready to plug into this equation. We’ll want to make one last change before we calculate 𝜆. Our kinetic energy is in units of electron volts, but we’d like to convert this to units of joules to be consistent with the rest of our values.

One electron volt is equal to 1.602 times 10 to the negative 19th joules. So if we multiply 2.0 times 10 to the sixth times 1.602 times 10 to the negative 19th, that gives us a value in joules equivalent to 2,000,000 electron volts, the given kinetic energy.

Using that value under our square root sign for the energy, when we compute 𝜆, we find that it’s equal to 20 times 10 to the negative 15th meters. This value is also known as 20 femtometers. That’s the de Broglie wavelength of a proton with this much kinetic energy.