00:01
For this exercise, we have to consider the poscen series of the hydrogen atom that takes an atom at an original energy level of ny, larger than 3, takes it down to the third energy level, so nf, the final energy level, is 3.
00:20
And in the first question, we're asked, what is the longest wavelength, lambda maximum, of this series? in order to calculate this, i got to remember that the energy of the energy level of the hydrogen atom is given by minus 13 .6 n squared electron volts.
00:44
And that from the conservation of energy, you have that the energy of the initial energy level is going to be the energy of the emitted photon, plus the energy of the final energy level.
00:58
Here, the final energy level is 3.
01:02
So you can calculate the energy of the photon as the energy of the final energy level, which is minus 13 .6 over n...
01:13
I'm sorry, the energy of the initial energy level.
01:16
It's minus 13 .6 over n .i squared, plus minus the energy of the third energy level, which is 13 .6 over 9.
01:34
So this is going to be 13 .619 minus 1 .i squared.
01:46
And notice that the energy of the photon equals hc over lambda, so that the maximum wavelength will happen when the energy is minimum.
02:01
And the minimum energy that can be achieved from this formula here will be achieved when n i is the smallest value possible.
02:13
And the smallest value possible for an atom to transition from some initial n to a final n equals 3 is n equals 4.
02:25
So we're going to have n y equals 4.
02:28
And we can calculate the smallest energy now...