### 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?

Weโre told in this statement that the box has a width 0.250 nanometers which weโll call ๐ฟ. We want to solve for the wavelength of the photons emitted when the electron transitions from the second excited state to the ground state. Letโs 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. Weโre told the electron starts at the second excited state where ๐ equals three, and it then transitions down to the ground state where ๐ is one. In the process, the electron emits a photon with a wavelength ๐. Itโs 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 ๐th energy level, ๐ธ sub ๐, equals ๐ squared times โ squared, where โ is Planckโs constant which weโll 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 ๐ times the width of the box ๐ฟ squared.

In our case, we have an electron that starts out at ๐ equals three and then moves to ๐ equals one. By the particle in a box energy equation, that equals nine โ squared over eight ๐๐ฟ squared minus one โ squared over eight ๐๐ฟ squared, which simplifies to โ squared over ๐๐ฟ squared. Now thatโs the change in energy of the electron, but what about the photon thatโs emitted? We can remember that photon energy equals โ times the frequency ๐ or โ times ๐ over ๐. All the energy of the electronโs transition is delivered to the emitted photon. That means we can write: โ squared over ๐๐ฟ squared is equal to the energy of the emitted photon โ ๐ over ๐. We can cancel out a factor of Planckโs constant from each side and then rearrange this equation to solve for ๐. And we find itโs equal to ๐, the mass of the electron, times ๐, the speed of light, times the width of the box, ๐ฟ squared, divided by โ, Planckโs constant.

In this example, weโll treat the mass of the electron ๐ as exactly 9.1 times 10 to the negative 31st kilograms and the speed of light ๐ as exactly 3.00 times 10 to the eighth meters per second.

When we plug in for ๐, ๐, ๐ฟ, and โ being careful to write our value for ๐ฟ in units of meters, when we calculate ๐, we find that, to three significant figures, it is 25.8 nanometers. That is the wavelength of the emitted photon.