Energy levels of Lithium. Quantum.

1. Dec 7, 2009

vwishndaetr

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,

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

Where the ground state, n=1,

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

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

So substituting,

$$e^2\rightarrow3e^2$$

So for Lithium,

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

But we know,

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

So,

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

With this we can conclude that,

$$E_t=\left(E_n+E_{n'}+E_{n''}\right)$$

So,

$$E_t=9\left(\frac{E_1}{n^2}+\frac{E_1}{n'\ ^2}+\frac{E_1}{n''\ ^2}\right)$$

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!

Last edited: Dec 8, 2009
2. Dec 8, 2009

weejee

$$E_t = 9\left(\frac{E_1}{n^2}+\frac{E_1}{{n^{\prime}}^2}+\frac{E_1}{{n^{\prime \prime}}^2}\right)$$

You have a nice formula for the energy. Why don't you evaluate it for different n, n' and n'' values and compare them with one another?

3. Dec 8, 2009

vwishndaetr

That is what I'm doing.

The states that are listed above, "1 1 2, 1 2 2, 1 1 3, 2 2 2, and 1 2 3", I plugged into that equation. The first number being n, second, n' and third being n''.

For example, 1 1 2, the ground state would be,

$$E_t = 9\left(\frac{E_1}{1^2}+\frac{E_1}{{1}^2}+\frac{E_1}{2^2}\right)= 9\left(\frac{E_1}{1}+\frac{E_1}{1}+\frac{E_1}{4}\right)= \frac{81E_1}{4}= \frac{81(-13.6\ eV)}{4}= -275.4\ eV$$

I understand how to use the formula.

Well let me ask this, do I need to use combinations of n, n', and n'' such that the energy levels increased minimally from one another?

What I think should be true is that E2 cannot be say 300eV if a combination of n's allows an energy of 295eV. So 295eV should be E2, not 300eV.

Get where I'm coming from? I just need to know if the last statements are true.

4. Dec 8, 2009

vwishndaetr

Bummer. I made a careless error. What a waste of time typing.

I, like a knuckle head did not actually try the states I suggested. I was looking at it as,

$$E = \left({n^2}+{{n^{\prime}}^2}+ {n^{\prime \prime}}^2}\right)$$

rather than,

$$E = \left(\frac{1}{n^2}+\frac{1}{{n^{\prime}}^2}+ \frac{1}{{n^{\prime \prime}}^2}\right)$$

I was careless. So 1 1 2, 1 1 3, 1 1 4, 1 1 5, and 1 1 6 are the correct ones.

Thanks for the help though.

5. Dec 8, 2009

RedX

I'm sort of confused. If you thought it was $$E = \left({n^2}+{{n^{\prime}}^2}+ {n^{\prime \prime}}^2}\right)$$, then wouldn't the lowest states be n=infinity, as then you'd get a huge negative number.

6. Dec 8, 2009

vwishndaetr

I hear ya on that part.

I was thinking the same thing, and am going to ask the good ol' professor tomorrow.