Question Video: Solving for Wavelength of Electrons Exiting Particle Accelerator Physics • Third Year of Secondary School

Video Transcript

A particle accelerator accelerates electrons through a potential difference Δ𝑉 equals 550 volts, as shown in the diagram. Find the wavelength of electrons as they exit the accelerator. Use a value of 1.60 times 10 to the negative 19 coulombs for the charge of electrons, a value of 9.11 times 10 to the negative 31 kilograms for the mass of electrons, and a value of 6.63 times 10 to the negative 34 joule-seconds for the Planck constant.

In our diagram, we imagine that this is the particle accelerator. An electron begins in the accelerator at this end and then is accelerated through a potential difference Δ𝑉 so that it leaves the accelerator here. The exiting electron has some velocity, and this corresponds to the wavelength of that electron. Here, we want to solve for that wavelength using the information given.

The first thing we can think about is this potential difference, Δ𝑉. In general, a potential difference — we’ll refer to it as Δ𝑉 — is equal to electrical potential energy, that’s what EPE stands for here, per unit charge, where 𝑞 is our charge. When our electron is at the far-right end of the accelerator, it has its maximum electrical potential energy. We can imagine this electron starting from rest then being accelerated across the potential difference Δ𝑉 and having all the electrical potential energy of the electron here being converted into kinetic energy of the electron here. In other words, across this accelerator, there’s an energy transfer that takes place. The original electrical potential energy of the electron becomes kinetic energy.

Beginning with our equation for Δ𝑉, let’s multiply both sides by the charge 𝑞, causing that factor to cancel on the right, which means the charge of our particle 𝑞 times Δ𝑉 is equal to the electric potential energy of that charge. We’ve just seen how the electric potential energy of our electrons is converted to kinetic energy. And recalling that kinetic energy in general equals one-half an object’s mass times its velocity squared, we can add to our equation so that we have 𝑞 times Δ𝑉 equals electric potential energy, which equals one-half 𝑚 times 𝑣 squared, where 𝑚 and 𝑣 refer to our exiting electrons.

We can now write this in a slightly simpler way by removing electrical potential energy from this line. We find that the charge of an electron 𝑞 times the potential difference Δ𝑉 through which that charge is accelerated is equal to one-half the mass of an electron 𝑚 times the exit velocity 𝑣 of that electron squared. Our problem statement asks us to solve for the wavelength of the exiting electrons.

Let’s recall that by the de Broglie relationship, the wavelength of an object is equal to the Planck constant divided by the mass of that object 𝑚 multiplied by its velocity. Notice that if we multiply both sides of this equation by 𝑣 divided by 𝜆, then on the left-hand side, the wavelength 𝜆 cancels out. And on the right, the velocity 𝑣 cancels. We find that velocity 𝑣 equals the Planck constant divided by 𝑚 times 𝜆. This relationship is helpful to us because we can use it to replace 𝑣 in this equation we generated earlier.

With this substitution made, we know that the Planck constant divided by 𝑚 times 𝜆 all squared is equal to the Planck constant squared divided by 𝑚 squared times 𝜆 squared. Note that one factor of the mass of the electron 𝑚 cancels from numerator and denominator. That gives us this equation. And recall that it’s the wavelength of 𝜆 that we want to solve for. To start doing that, let’s multiply both sides of the equation by 𝜆 squared, canceling 𝜆 squared on the right, and then multiply both sides by one divided by 𝑞 times Δ𝑉. When we do this, on the left, the 𝑞 in numerator and denominator cancel as do the Δ𝑉 in numerator and denominator.

We find that 𝜆 squared equals one-half times the Planck constant squared divided by 𝑚 times 𝑞 times Δ𝑉. And then lastly, taking the square root of both sides, on the left, the square root and the square of 𝜆 cancel one another out, while on the right-hand side the equation simplifies to the Planck constant divided by the square root of two times 𝑚 times 𝑞 times Δ𝑉.

In our scenario, 𝑞 is 1.60 times 10 to the negative 19 coulombs, the charge of an electron. 𝑚 is 9.11 times 10 to the negative 31 kilograms, the mass of an electron. And Δ𝑉, we’re given, is 550 volts. Knowing this and having a value to use for the Planck constant, let’s clear some space on screen and move ahead with substituting in the values that appear on the right-hand side of this equation.

We have the Planck constant in the numerator, the mass of the electron, the charge of the electron, and the potential difference in the square root of the denominator. And when we calculate the result, we find an answer of 5.2359 and so on times 10 to the negative 11 meters. We can choose to write our answer though to one decimal place. Rounding our result to this level precision, we have 5.2 times 10 to the negative 11 meters. This is the wavelength of the electrons as they exit the accelerator.