- #1

fluidistic

Gold Member

- 3,947

- 263

## Homework Statement

Find all the terms of an atom whose last subshell is np³.

## Homework Equations

[itex]M_L=\sum _i m_{l _i}[/itex]

[itex]M_S=\sum _i m_{s _i}[/itex]

## The Attempt at a Solution

My professor did the same exercise but with np². Basically he wrotes all the possible quantum numbers for the atom:

1)[itex]m_l=1[/itex], [itex]m_s=1/2[/itex]. 2)[itex]m_l=0[/itex], [itex]m_s=1/2[/itex]. 3)[itex]m_l=-1[/itex], [itex]m_s=1/2[/itex]

1')[itex]m_l=1[/itex], [itex]m_s=-1/2[/itex]. 2')[itex]m_l=0[/itex], [itex]m_s=-1/2[/itex]. 3')[itex]m_l=-1[/itex], [itex]m_s=-1/2[/itex]

Then he calculated all the possible values for [itex]M_L[/itex] and [itex]M_S[/itex]. There are 15 values in total.

After this, there is an obscure step to me (he counted I don't really know what) and went to the conclusion that the solution to the problem is [itex]^1 D ^3 P ^1 S[/itex]. Where the upper script is worth [itex]2S+1[/itex].

So I did the same method as him for np³ (I guess this notation means that there are 3 electrons on the subshell p or an arbitrary n?). I got 20 values for [itex]M_L[/itex], [itex]M_S[/itex]. I'm stuck at doing the obscure step now. I have all possible values for [itex]M_L[/itex] and [itex]M_S[/itex].

Can someone explain me what I should do next?

Another question is... since n seems arbitrary, can I for example take [itex]n=1[/itex], so that [itex]l=0[/itex] and [itex]m_l=0[/itex]. My professor seems to have taken n=2 for some misterious reason to me. Does someone understand why?

Edit: since there are at least 3 electrons I guess I cannot take n=1, since at least n=2. Ah... n must equal 2... ok that's what I considered, good. So I'm stuck where I pointed out.

Last edited: