### Video Transcript

An electron is confined to a box of width 0.250 nanometres. What is the wavelength of photons emitted when the electron transitions between the third and second excited states?

This statement tells us the width of the box the electron is confined to; itβs 0.250 nanometres. Weβll call that value πΏ. We want to solve for the wavelength of photons emitted when the electron in the box transitions from the third to the second excited states. To begin our solution, letβs draw a diagram of this scenario.

We have a box β an infinite potential well β of width πΏ and weβre told that in this box an electron starting out at the third excited energy level, where π equals four, then transitions down to the second excited level, where π equals three. In the process of this transition, the electron emits a photon, a particle of light with wavelength π.

We can start solving for that wavelength by remembering that for a particle in a box like the electron is here, the πth energy level of that particle πΈ sub π ist equal to π, an integer, squared times β, Planckβs constant, squared divided by eight times the electron mass π times the width of the box πΏ squared. Regarding this equation, weβll assume that β is exactly 6.626 times 10 to the negative 34th joule seconds and that π, the mass of the electron, is 9.1 times 10 to the negative 31st kilograms.

When you apply this relationship to the transition in our problem statement, we see that the change in energy of the electron ΞπΈ equals the energy at the third excited level, where π equals four, minus the electron energy at the second excited level, where π equals three. This is equal to 16 β squared over eight ππΏ squared minus nine β squared over β [eight] ππΏ squared or seven β squared over eight ππΏ squared. Thatβs the change in energy of the electron in the well.

That change in energy is equal to the energy of the photon thatβs emitted during that transition. Photon energy πΈ is equal to β times the frequency π or β times π over π, where π will treat as exactly 3.00 times 10 to the eighth meter per second. So photon energy β π over π equals seven β squared over eight ππΏ squared. A factor of Planckβs constant cancels. And when we rearrange to solve for π, we find that itβs eight πππΏ squared over seven times β.

We can now plug in for π, π, πΏ, and β. When we plug those values in, being careful to use a value in units of metres for the width of the well πΏ and then enter them on our calculator, we find that to three significant figures the emitted photon has a wavelength of 29.4 nanometres.