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vwishndaetr

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I am having an issue calculating the first few energy levels of Lithium. To add, electron-electron repulsions are ignored.

Since electron-electron repulsion is being ignored, I am starting with the following,

For Hydrogen,

[tex]E_n=-\left[\frac{m}{2\hbar^2}\left(\frac{e^2}{4\pi\epsilon_0}\right)^2\right]\frac{1}{n^2}=\frac{E_1}{n^2}[/tex]

Where the ground state, n=1,

[tex]E_1=-\left[\frac{m}{2\hbar^2}\left(\frac{e^2}{4\pi\epsilon_0}\right)^2\right]=-13.6eV[/tex]

As mentioned, since we are ignoring electron-electron repulsion, the only difference will the amount of charge.

So substituting,

[tex]e^2\rightarrow3e^2[/tex]

So for Lithium,

[tex]E_n=-\left[\frac{m}{2\hbar^2}\left(\frac{3e^2}{4\pi\epsilon_0}\right)^2\right]\frac{1}{n^2}=-9\left[\frac{m}{2\hbar^2}\left(\frac{e^2}{4\pi\epsilon_0}\right)^2\right]\frac{1}{n^2}[/tex]

But we know,

[tex]E_n=-\left[\frac{m}{2\hbar^2}\left(\frac{e^2}{4\pi\epsilon_0}\right)^2\right]\frac{1}{n^2}=\frac{E_1}{n^2}[/tex]

So,

[tex]E_n=-9\left[\frac{m}{2\hbar^2}\left(\frac{e^2}{4\pi\epsilon_0}\right)^2\right]\frac{1}{n^2}=9\frac{E_1}{n^2}[/tex]

With this we can conclude that,

[tex]E_t=\left(E_n+E_{n'}+E_{n''}\right)[/tex]

So,

[tex]E_t=9\left(\frac{E_1}{n^2}+\frac{E_1}{n'\ ^2}+\frac{E_1}{n''\ ^2}\right)[/tex]

Now here is where my question comes in. I need to find first five energy levels of Lithium. I am a little confused on to which values of n, n', and n'' to choose.

A friend of mine said that first five states would be, 1 1 2, 1 1 3, 1 1 4, 1 1 5, and 1 1 6.

I on the other hand think it should be, 1 1 2, 1 2 2, 1 1 3, 2 2 2, and 1 2 3.

Can someone shed some light? I always thought to go by the sums of the squares of each n combination. So my order follows chronologically from lowest to the next possible energy level. His on the other hand, seem to skip one's that I think should be present.

All help is appreciated!

Since electron-electron repulsion is being ignored, I am starting with the following,

For Hydrogen,

[tex]E_n=-\left[\frac{m}{2\hbar^2}\left(\frac{e^2}{4\pi\epsilon_0}\right)^2\right]\frac{1}{n^2}=\frac{E_1}{n^2}[/tex]

Where the ground state, n=1,

[tex]E_1=-\left[\frac{m}{2\hbar^2}\left(\frac{e^2}{4\pi\epsilon_0}\right)^2\right]=-13.6eV[/tex]

As mentioned, since we are ignoring electron-electron repulsion, the only difference will the amount of charge.

So substituting,

[tex]e^2\rightarrow3e^2[/tex]

So for Lithium,

[tex]E_n=-\left[\frac{m}{2\hbar^2}\left(\frac{3e^2}{4\pi\epsilon_0}\right)^2\right]\frac{1}{n^2}=-9\left[\frac{m}{2\hbar^2}\left(\frac{e^2}{4\pi\epsilon_0}\right)^2\right]\frac{1}{n^2}[/tex]

But we know,

[tex]E_n=-\left[\frac{m}{2\hbar^2}\left(\frac{e^2}{4\pi\epsilon_0}\right)^2\right]\frac{1}{n^2}=\frac{E_1}{n^2}[/tex]

So,

[tex]E_n=-9\left[\frac{m}{2\hbar^2}\left(\frac{e^2}{4\pi\epsilon_0}\right)^2\right]\frac{1}{n^2}=9\frac{E_1}{n^2}[/tex]

With this we can conclude that,

[tex]E_t=\left(E_n+E_{n'}+E_{n''}\right)[/tex]

So,

[tex]E_t=9\left(\frac{E_1}{n^2}+\frac{E_1}{n'\ ^2}+\frac{E_1}{n''\ ^2}\right)[/tex]

Now here is where my question comes in. I need to find first five energy levels of Lithium. I am a little confused on to which values of n, n', and n'' to choose.

A friend of mine said that first five states would be, 1 1 2, 1 1 3, 1 1 4, 1 1 5, and 1 1 6.

I on the other hand think it should be, 1 1 2, 1 2 2, 1 1 3, 2 2 2, and 1 2 3.

Can someone shed some light? I always thought to go by the sums of the squares of each n combination. So my order follows chronologically from lowest to the next possible energy level. His on the other hand, seem to skip one's that I think should be present.

All help is appreciated!

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