# Determining ionization energy of Lithium.

brbrett

## Homework Statement

What is the third ionization energy of Li in its ground state?
A. 4.91 E-18 J
B. 6.54 E-18 J
C. 7.79 E-18 J
D. 9.20 E-18 J
E. 1.96 E-17 J

## Homework Equations

Maybe:
EN=-(2.18E-18)(Z2/N2)

## The Attempt at a Solution

Plugging in values doesn't give me any of the answers. At first I thought Z was the number of electrons, but now I think it is the atomic number. Either way, I'm at a loss for this question.

Mentor
How many electrons does lithium have if two were removed already? Where are the remaining electrons?
At first I thought Z was the number of electrons, but now I think it is the atomic number.
Quantities used in formulas are explained at the place where you found the formula. There is no need to guess.

brbrett
Lithium should have 1 electron remaining if two were removed. Also, we weren't told in the textbook I'm using what Z is. We were told about every other value though, which I find is funny.

Back on topic though, I did think of this, but I'm not sure how I can get a number out of this. Wouldn't the ionization energy of the lone electron be much greater considering how two have already been removed? How can I account for this analytically?

Thanks!

Mentor
Lithium should have 1 electron remaining if two were removed.
Right.
Also, we weren't told in the textbook I'm using what Z is.
It would surprise me if the textbook doesn't mention that.
It is the charge the electron "sees", in this case simply the charge of the nucleus, also called atomic number.
Wouldn't the ionization energy of the lone electron be much greater considering how two have already been removed?
Sure, the first and second ionization energies are lower.
How can I account for this analytically?
Reduce the charge of the nucleus by the charge of all "inner" electrons. That is not exact, but it gives a reasonable approximation for most atoms.

• brbrett
brbrett
Ok, so now I vaguely recall the concept of Effective charge. If I were to plug this into my formula, would I want to replace (Z2/N2) with Δ(Z2/N2)?

To clarify as to what I would plug in:
E=-(2.18E-18)((0/1)-(3/1))=6.54E-18
This is one of the given answers, which is good, however, the answer is E (I wasn't confident in this answer anyways).

For 0/1, I made Z=0 as I would suppose that, as there are no electrons remaining, there is not charge being applied to them. I used N=1 because that is the shell we are dealing with.

Getting closer, but still a wee bit lost. An interesting question, though. Thanks for the help!

Mentor
Ok, so now I vaguely recall the concept of Effective charge. If I were to plug this into my formula, would I want to replace (Z2/N2) with Δ(Z2/N2)?
I'm not sure what you mean by the second expression.
To clarify as to what I would plug in:
E=-(2.18E-18)((0/1)-(3/1))=6.54E-18
This is one of the given answers, which is good, however, the answer is E (I wasn't confident in this answer anyways).
Don't forget to square Z.

There is no need to take a difference. The formula gives the energy relative to free electrons, which is exactly the final state you want, so you can just plug in Z and N and you get the ionization energy.

Mentor
Wouldn't hurt to state the obvious - twice ionized lithium is a hydrogen-like atom (single electron and a nucleus), so finding the ionization energy is just a direct application of the Rydberg formula. No need for effective charge nor any other complications.

brbrett
Thanks! I'm starting to get it now. One question I have is why would 32/1 be negative? I know it must be, considering how the answers in the MC are all positive, and there is a negative sign in front of the constant.

Also all the talk about effective charge was good review for me, so it helps anyways.

Mentor
The ionization energy is positive by definition. It is the energy you have to add to the electron to set it free. As its potential energy afterwards is zero (again by definition), it means the energy before was negative.