The so-called Lyman $-\alpha$ photon is the lowest energy photon in the Lyman series of hydrogen and results from an electron transitioning from the n 5 2 to the n 5 1 energy level. Determine (a) the energy in eV and (b) the wavelength in nm of a Lyman - a photon.
(a) $ E = 10.2 \, \rm eV $
(b) $ \lambda = 121.6 \, \rm nm $
this exercise, we have to calculate the energy and the wavelength off the Lyman Alfa Fulton that's emitted when the hydrogen atom transitions from the second interview level to the 1st 1 So the first thing that we need to remember is that the energy of the end energy level in the hydrogen atom is minus 13.6, divided by n squared. Okay, and also from conservation of energy. We have that the energy of the emitted Fulton is given by the energy of the initial energy level, which which in this case is to mine, is the energy of the final energy level, which is you want in this case. So this is equal to 13.6 times 1/1 squared. That's just one minus 1/2 squared, which is 1/4. This is an electron volts, of course, because ah, the energy is 13.6 electoral votes divided by and square. So this is 13.6 times 3/4 electoral votes. So we have the energy of a photon is equal to 10.2 electron volt. That's the energy of the well. Im on Alfa folded and we can calculate the wavelength of Fulton, remembering that the energy is equal to age see over longer. So Lunda is a seaworthy energy, A C stuff. 100 and 40 electoral votes and the meters and the energy is 10.2 electoral votes. So Lunda is 101 121.6 centimeters. Okay, so