Stern-Gerlach Predictions of Elements

[NeonCannon]

I am baffled at the answers to the following HW quesitons compared with what I thought I knew about the topic. Hoping I can get my misunderstandings cleared up.
Question 1
How many lines are expected on a detector plate of a Stern-Gerlach experiment if a beam of
a) potassium atoms
b) calcium atoms
c) oxygen atoms
d) tin atoms

Answers
1a) 2
1b) 1
1c) 5
1d) 1

Solution Attempt
The Stern-Gerlach experiment puts atoms through an inhomogeneous magnetic field. A magnetic moment in a magnetic field will experience a force. The magnitude of the force is proportional to the magnetic moment, and the magnetic moment is proportional to the angular momentum. The variation in the force yields the varying number of bands.

The spin of an electron contributes to the total angular momentum of the electron. Therefore, I should consider the atoms electron configuration somehow.

1a) Potassium has filled in the 1s2s2p3s3p orbitals. The electrons here do not contribute at all to the overall angular momentum. Only its single electron in the 4s state. The 4s orbital corresponds to l=0 which provides only a single magnetic moment state (m=0). The electron is unpaired, therefore it is free to obtain an up or down spin. Thus, only two values of its total angular momentum are possible and two bands are expected.

1b) Calcium is similar to Potassium except it has two electrons in its 4s. The up spin and down spin are paired and cancel each other out. Finally, since there is only a single magnetic moment and no spin: one band is expected.

1c) Oxygen fills in the 1s2s orbitals. These electrons do not contribute to the angular momentum. The only electrons which matter are the 4 electrons in 2p. The p orbital refers to l=1 and therefore m=-1, 0, 1 for the magnetic moment are possible. Since there is an even number of electrons they are paired (up with down spin) and no angular momentum is contributed from the spin. I would expect only 3 bands in the experiment.

1d) With similar reasoning in 1a, 1b, 1c, the only electrons which matter in Tin atoms are the 4 electrons in the 5p orbital. The p orbital refers to l=1 and gives magnetic moments of m=1, 0, -1 and the paired electrons give no spin. So, I would expect only 3 bands.

**As I was told the answers, I do not understand what is really wrong with my reasoning.

 

[M Quack]

1a correct.

1b correct.

1c using Hund's rules, you first couple the orbital momentum of the p electrons. you get L=1

Next, you couple the spins. You get S=1.

Then you couple L and S. Since the shell is more than half filled, you get J=L+S=2. There are 2J+1 lines.

1d. You have 2 p-electrons. First you couple L=1, next you couple S=1. The shell is less than half filled, so J=L-S=0. Just one line.

 

[Jonesgreg614]

Hey M Quack - Thanks for explaining it out. However, I don't see how this would work for an element like Phosphorus. Could you perhaps run the coupling numbers for that element? I'm getting fractions and I don't know how you could get fractions of lines from the SG experiment. Your help is greatly appreciated :)

 

[M Quack]

http://www.webelements.com/phosphorus/atoms.html

The electronic configuration is [Ne] 3s2 3p3.

3s is full, so no magnetic moment

3p has 3 electrons, according to Hund's rules all same spin and one each for m=-1, 0, 1.

that makes L=0 and S=3/2 so J = L+S =3/2

This gives 2J+1 = 4 lines

How do you manage to get fractions?

 

[Jonesgreg614]

Hey M Quack

Thanks for your quick reply! I had Hund's rules down but for some strange reason I was missing the (2) in front for the J, which was giving me fractions. Weird.

I'm able to use Hund's law to determine these lines and I see that an atom like K or Aq will produce two lines (each with a 50% occurrence) - What percentage is attributed to each of the four lines of Phosphorus? Am I missing someway to calculate this or is it a simple, even distribution? Thanks so much M Quack!

 

[M Quack]

I would guess the 4 lines have pretty much equal intensity, but I am not an expert.

In a magnetic field each state has a different energy. But the difference in energy is probably much smaller than kT.