Calculate the wavelength in nm of the light emitted when an electron makes a transition from an orbital in n = 5 to an orbital in n = 2 in the hydrogen atom. (LO 5.8) (a) 2.31 * 10-3 nm (b) 4.34 * 10-2 nm (c) 231 nm (d) 434 nm

Video transcript

Hey everyone were asked what is the wavelength of light and nanometers emitted when an electron in a hydrogen atom transitions from an orbital in an equal six to n equals three. To answer this question, we need to use our Balmer Rydberg equation, which is one over wavelength equals our Rydberg constant times one over n f squared -1 over an initial squared. And our Rydberg constant is going to be our equals 1. times 10 to the negative two nanometers to the negative first. And for this example, our N final is going to be three and our N initial is going to be six since our hydrogen atom transitions from an equal six to N equals three. So, plugging in our values, we get one over wavelength equals 1.97 times 10 to the negative two nanometers. The negative first which is our ride for constant times one over NF squared, which will be three squared minus one over an initial squared which will be six squared. So when we calculate this out, we get a one over wavelength of 9. times 10 to the negative four nanometers to the negative first. To calculate for our wavelength, we end up with a wavelength of 1,093 nm which is going to be our final answer. Now, I hope that made sense. And let us know if you have any questions