# Beryllium atom in a weak magnetic field

1. Aug 4, 2012

### fluidistic

1. The problem statement, all variables and given/known data and attempt at a solution
Considering that the ground state of the beryllium atom is $2s^2$ and the first excited state is $2s2p$, I was wondering in how many lines would the emission/absorption spectrum be modified in a weak magnetic field.
I know that if there's no magnetic field, only "1" frequency/wavelength will be emitted/absorbed.
In order for a weak magnetic field to change this, there would need to have the denegeration of the quantum numbers $m_j$'s to be eradicated. This is so because in the presence of a weak magnetic field, $\Delta E = \mu _B g_L m_j B_{\text {weak}}$.

So for the ground state, the outer 2 electrons are in a subshell that can contain at most 2 electrons, therefore it is filled entirely. Thus in this case $J=|l+s|$ where $l=0$ and s=0. s=0 because $s=\sum m_s =1/2-1/2$ thanks to Pauli exclusion's principle. Thus in this case $J=0$, hence $m_j=0$. The ground state won't be affected by any weak magnetic field.

Now for the first excited state, 2s2p. There's 1 electron in the shubshell 2s, half filled. So that $J=|l-s|$ where $l=0$ and s=1/2. This makes $J=1/2$.
For the outer electron, it is in a subshell that can contains up to 6 electrons, thus again $J=|l-s|$, where this time $l=1$ and s=1/2. This makes $J=1/2$. So that the total angular momentum $J=1/2+1/2=1$. This would make $m_j$ going from -1 to 1, passing by 0. In this case it seems that the first excited state gets splitted into 3 energy levels (one of which was the one without any magnetic field) when we apply a weak magnetic field.
So that all in all there would be a 3 spetrum "lines" or frequencies/wavelengths if we apply a weak B field (let's say 1 tesla at most).

However my friend told me that there's no splitting at all when there's a weak magnetic field. I wonder where I went wrong...

2. Aug 7, 2012

### fluidistic

Do you guys have any idea on my "error"?