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fluidistic

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**1. Homework Statement and attempt at a solution**

Considering that the ground state of the beryllium atom is [itex]2s^2[/itex] and the first excited state is [itex]2s2p[/itex], 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 [itex]m_j[/itex]'s to be eradicated. This is so because in the presence of a weak magnetic field, [itex]\Delta E = \mu _B g_L m_j B_{\text {weak}}[/itex].

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 [itex]J=|l+s|[/itex] where [itex]l=0[/itex] and s=0. s=0 because [itex]s=\sum m_s =1/2-1/2[/itex] thanks to Pauli exclusion's principle. Thus in this case [itex]J=0[/itex], hence [itex]m_j=0[/itex]. 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 [itex]J=|l-s|[/itex] where [itex]l=0[/itex] and s=1/2. This makes [itex]J=1/2[/itex].

For the outer electron, it is in a subshell that can contains up to 6 electrons, thus again [itex]J=|l-s|[/itex], where this time [itex]l=1[/itex] and s=1/2. This makes [itex]J=1/2[/itex]. So that the total angular momentum [itex]J=1/2+1/2=1[/itex]. This would make [itex]m_j[/itex] 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...