00:01
So part here, this problem asks for the energy of a photon emitted from the 3d to the 2p state.
00:06
So we know the formula e equals hc over lambda, where h is planks constants given an electron volt seconds.
00:14
Lambda is the way of length given in the problem and sees the speed of light.
00:17
And if you plug in those values, you'll get 2 .612 electron volts.
00:23
And then part b of the problem is asking us to use selection rules, describe section 41 .4.
00:32
To find the allowed transitions.
00:35
So based on the selection rules, we know that delta l is 1, and correspondingly delta ml can be basically negative l, 0 to positive l, so 0 plus or minus 1.
00:49
So if we write out all the states for, if we write out the quantum numbers for 3d, states 3d and 2p for 3d, n equals 3l equals 2, and ml is plus or minus, 2 plus or minus 1 and 0 for 2 p, n equals 2 l equals 1 and then m l is 0 and plus or minus 1 and so listed on the left here are all the possible combinations where l is equal to 2 and m l is equal to 2 2 1 0 negative 1 or negative 2 and so we want the transition to go with where l jumps down 1 or ml just not jump down at all or goes plus or minus 1.
01:47
So listed are just all the combinations where l goes, where l prime minus l is 1, so that's 2 for l and 1 for l prime and 1 for l prime and where ml is 0, 0, 1 or negative 1...