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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 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...
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...