Determining ionization energy of Lithium.

AI Thread Summary
The discussion revolves around calculating the third ionization energy of lithium in its ground state. Participants clarify that the atomic number (Z) represents the charge of the nucleus, and after two electrons are removed, lithium has one electron remaining. The formula for ionization energy is applied, with emphasis on correctly using Z and N values, leading to the conclusion that lithium behaves like a hydrogen-like atom after two ionizations. The ionization energy must be positive, as it represents the energy required to free the electron, which has negative potential energy before ionization. Overall, the conversation highlights the importance of understanding effective charge and the application of the Rydberg formula in this context.
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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.
 
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How many electrons does lithium have if two were removed already? Where are the remaining electrons?
brbrett said:
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.
 
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!
 
brbrett said:
Lithium should have 1 electron remaining if two were removed.
Right.
brbrett said:
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.
brbrett said:
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.
brbrett said:
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.
 
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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!
 
brbrett said:
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.
 
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.
 
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.
 
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.
 
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