# Energy levels of Lithium. Quantum.

• vwishndaetr
In summary, Lithium has a ground state of 1, with the next five energy levels being 1 1 2, 1 1 3, 1 1 4, 1 1 5, and 1 1 6.
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:
$$E_t = 9\left(\frac{E_1}{n^2}+\frac{E_1}{{n^{\prime}}^2}+\frac{E_1}{{n^{\prime \prime}}^2}\right)$$

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?

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

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?

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.

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.

vwishndaetr said:
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.

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.

I hear you on that part.

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

## What is the energy level of lithium?

The energy level of lithium is determined by the number of electrons in its outer shell. It has three energy levels, with two electrons in the first level, and one electron in the second level.

## How are energy levels of lithium related to quantum mechanics?

The energy levels of lithium are related to quantum mechanics because they are determined by the quantized energy states of the electrons in the atom. Quantum mechanics describes the behavior of particles on a very small scale, such as atoms and subatomic particles.

## How do energy levels of lithium affect its chemical properties?

The energy levels of lithium play a crucial role in its chemical properties. The electrons in the outer energy level are involved in chemical reactions, and the number of electrons in this level determines how lithium will interact with other atoms.

## Why does lithium have fewer energy levels than other elements?

Lithium has fewer energy levels than other elements because it has a smaller number of electrons in its outer shell. As a result, it requires less energy to fill its outer energy level and achieve a stable electron configuration.

## How are the energy levels of lithium observed and measured?

The energy levels of lithium can be observed and measured through various spectroscopic techniques, such as atomic absorption spectroscopy and emission spectroscopy. These methods use light to excite the electrons in the atom and measure the energy levels based on the emitted or absorbed wavelengths of light.

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