00:01
So the essence of this problem is how different isotopes have different wavelengths in their emission spectra.
00:09
And it's a reduced mass formula.
00:12
So what's happening is that when the reduced mass goes up, the energy spacing goes up, and the wavelength of the emitted photon goes down.
00:33
So a heavier isotope is going to get.
00:35
You a shorter wavelength than the emissions.
00:40
And we want to know, particularly hydrogen, the wavelength difference between deuterium and hydrogen one.
00:56
So the way we approach this is by taking the reduced mass energy equation.
01:06
We'll do hydrogen one first.
01:13
The electronic structure is the same in the two situations.
01:17
It's only the mass that's different.
01:18
So in both cases, you have one proton, one electron.
01:24
So the charge is e, q1 and q2 squared, and q2 squared.
01:30
So it's e to the fourth power is the charge term in all cases.
01:36
So this, this, uh, parentheses term is going to be the same in all situations.
01:44
And we're going to be able to factor it out.
01:53
So to simplify things, i'll just define a constant is equal to k squared, e of the fourth divided by 2h bar, where k is kulam's constant, and e is the electric charge.
02:11
And that works out to be 2 .394 times 10 to the 12 meter squared per second squared.
02:25
So it follows from that that in an energy transition, the initial state to the final state, and then these minus signs reverse the order, and then the change in energy ends up being positive.
03:46
This is dropping a level, so one over the final n squared is a larger number.
03:55
So then to know the wavelength, the formula is based on the energy equals plogs constant times the frequency.
04:25
It's just rearranged.
04:31
So we can substitute in the expression we just did for the change in energy in the denominator.
05:04
So for that was for hydrogen 1 for deuterium, it's the exact same logic except for the reduced match mu -sub -d.
05:41
You follow the exact same...