An energetically excited hydrogen atom has its electron in a 5f subshell. The electron drops down to the 3d subshell, releasing a photon in the process. (b) What wavelength of light is emitted by the process?

Video transcript

Hello everyone today. We're at the falling problem calculate the wavelength of light emitted when an electron of a hydrogen atom drops from four D 23 P. So first we need to recall our board equation formula for emission. It's gonna be one over the wavelength is equal to R. Which is gonna be a constant called the Rydberg constant. And we're gonna multiply that constant by one, divided by our In squared final. And that's going to be basically the quantum number -1, divided by our quantum number squared again. But for the initial. So it was stated that the electron in the hydrogen atom dropped from 40 to 3 P. And so only the principal quantum number N is considered for the wavelength due to the hydrogen atom of a particular energy level having all the same energy. So we would call that degenerate. But for this problem sake we will go ahead and continue working through it. So ride Berg's constant is 1.09 seven times 10 to the 7th. Power m raised to the power of -1 and then we're going to have one divided by. So it dropped from 4 to 3. So our final energy level is three. We're gonna square that. Then one minus it came from the initial which was four. We're gonna square that. Once again we're still going to have our one over the wavelength. However, will we simplify this? We will get 533,263 m. Race of the power of one. If we were to simplify this, this would be our wavelength being equal to one divided by our our 533,263 m raised to the negative one, which when we solved this would give us a final answer of a wavelength equaling 1.88 times 10 to the negative six m. And so with that we have answered the question overall, I hope this helped, and until next time.