Stern Gerlach confusion

In this experiment, when hydrogen atoms are projected through the inhomogeneous magnetic field we get two distinct 'traces' on the screen because of the spin-up/spin-down possibilties.

But I'm struggling to extend this concept to atoms with more optically active electrons. Say we have helium atoms (2 o.a. electrons) being shot through the magnetic field - how do I determine the number of traces then? I understand that for a configuration with a spin S, there are 2S + 1 possibilties. So do I then consider the total spin - which can either be 0 or 1 in this case - and assume that we can have 2*1 + 1 = 3 possibilties = 3 traces for helium?

Thanks for any help.

Reply 1

spread_logic_not_hate

Original post by trm90

I would agree with you and say you consider the total electron spin, as effectively that is what you're doing in the case of hydrogen as well (total spin S =

\dfrac{1}{2}

, hence there are 2 possible states)

That makes sense for helium as well as you could have the electron states (u being spin up, d being spin down):

u u (S = 1, Sz = 1) [bends 'upwards' through Stern-Gerlach equipment]
u d (S = 0, Sz = 0) [Straight through, no deviation]
d u (S = 0, Sz = 0) [Straight through, no deviation]
d d (S = 1, Sz = -1) [bends 'downwards' through Stern-Gerlach equipment]

i.e. three distinct paths (as the u d spin state is indistinguishable from the d u spin state)

(edited 13 years ago)

Reply 2

suneilr

Helium in the ground state is S=0 iirc so it should only have one trace. For excited states there should be either 1 or 3 traces depending on whether para or orthohelium is used.

Reply 3

suneilr

Original post by spread_logic_not_hate

I would agree with you and say you consider the total electron spin, as effectively that is what you're doing in the case of hydrogen as well (total spin S =

\dfrac{1}{2}

I'm pretty sure ud and du aren't physical states. You need linear combinations of those states ie ud+du and ud-du. So you end up with three S=1, and one S = 0 possible states depending on the symmetry of the spatial wavefunction.

Reply 4

trm90

Original post by suneilr

Helium in the ground state is S=0 iirc so it should only have one trace. For excited states there should be either 1 or 3 traces depending on whether para or orthohelium is used.

Ah right, so we assume it's parahelium... thanks :smile:

One more question. Now say I want to extend this trace problem to Boron, which has three optically active electrons. I'm told, as a hint, that Boron has two L-S coupling states. The electrons are in the second shell so that n = 2 and l = 1. I'm assuming the two L-S coupled states are basically either J = L + S or J = |L - S|, where L and S are the total angular momenta and spins, respectively.

But I'm still lost. No idea if I'm missing out on some 'rule' that governs how many traces there will be :frown:

Reply 5

suneilr

Original post by trm90

Ah right, so we assume it's parahelium... thanks :smile:

I'm pretty sure that the number of traces is given by the number of m_j values. Therefore the number of traces will depend on which LS coupling state the boron is in.

Reply 6

trm90

suneilr

Right, so as I vaguely remember there are 2J + 1 m_J values as

-J \leq m_J \leq +J

. That means J = L + S = 3 + 3/2 = 9/2, or J = L - S = 3 - 3/2 = 3/2. Which gives 2(9/2) + 1 = 10 traces or 2(3/2) + 1 = 4 traces.

So a total of 14 traces then?

Reply 7

suneilr

Original post by trm90

Right, so as I vaguely remember there are 2J + 1 m_J values as

-J \leq m_J \leq +J

. That means J = L + S = 3 + 3/2 = 9/2, or J = L - S = 3 - 3/2 = 3/2. Which gives 2(9/2) + 1 = 10 traces or 2(3/2) + 1 = 4 traces.

So a total of 14 traces then?

I think you've got your J values wrong. You've got electrons in n=2, l= 0 so these don't contribute to L. They also have to have opposite spin and and so S=0. So only the n=2, l=1 electron contributes, and since theres only one electron you can take L=1, and then the spin can be either aligned or antialigned with L, so you have J = L +S = 1 + 1/2 = 3/2 or J=L-S = 1-1/2 = 1/2 so you end up with either 2 or 4 traces depending on the LS state.

(edited 13 years ago)

Reply 8

trm90

Original post by suneilr

I think you've got your J values wrong. You've got electrons in n=2, l= 0 so these don't contribute to L. They also have to have opposite spin and and so S=0. So only the n=2, l=1 electron contributes, and since theres only one electron you can take L=1, and then the spin can be either aligned or antialigned with L, so you have J = L =S = 1 + 1/2 = 3/2 or J=L-S = 1-1/2 = 1/2 so you end up with either 2 or 4 traces depending on the LS state.

Ohhhhhhhhhhhhhhhhhhhhhhhhh.

Damn, this is so embarrassing. I've been counting my electrons wrong: for some reason I thought I had to put 2 electrons in the first shell, then up to 8 in the subsequent shell and just didn't count it properly at all. I understand now that there is only one L-S coupled electron in the n = 2 l = 1 state.

Thanks again so much suneil, you're a star. And sorry for my denser-than-usual posts these days >_<