### Video Transcript

An electron is confined to a box of width 0.250 nanometers. What is the wavelength of photons emitted when the electron transitions between the second excited state and the ground state?

Weβre told in this statement that the box has a width 0.250 nanometers which weβll call πΏ. We want to solve for the wavelength of the photons emitted when the electron transitions from the second excited state to the ground state. Letβs start our solution by drawing a diagram. In this example, we have an infinitely deep box the electron is not able to escape out of. Weβre told the electron starts at the second excited state where π equals three, and it then transitions down to the ground state where π is one. In the process, the electron emits a photon with a wavelength π. Itβs that wavelength we want to solve for.

To do that, we can recall that for a particle in a box, like we have here, the energy of the particle at the πth energy level, πΈ sub π, equals π squared times β squared, where β is Planckβs constant which weβll take to equal exactly 6.626 times 10 to the negative 34th joule seconds, all of which is divided by eight times the mass of the electron π times the width of the box πΏ squared.

In our case, we have an electron that starts out at π equals three and then moves to π equals one. By the particle in a box energy equation, that equals nine β squared over eight ππΏ squared minus one β squared over eight ππΏ squared, which simplifies to β squared over ππΏ squared. Now thatβs the change in energy of the electron, but what about the photon thatβs emitted? We can remember that photon energy equals β times the frequency π or β times π over π. All the energy of the electronβs transition is delivered to the emitted photon. That means we can write: β squared over ππΏ squared is equal to the energy of the emitted photon β π over π. We can cancel out a factor of Planckβs constant from each side and then rearrange this equation to solve for π. And we find itβs equal to π, the mass of the electron, times π, the speed of light, times the width of the box, πΏ squared, divided by β, Planckβs constant.

In this example, weβll treat the mass of the electron π as exactly 9.1 times 10 to the negative 31st kilograms and the speed of light π as exactly 3.00 times 10 to the eighth meters per second.

When we plug in for π, π, πΏ, and β being careful to write our value for πΏ in units of meters, when we calculate π, we find that, to three significant figures, it is 25.8 nanometers. That is the wavelength of the emitted photon.