# Determining ionization energy of Lithium.

• brbrett
In summary, ionization energy is the amount of energy needed to remove an electron from an atom or molecule in its gaseous state. It is an important factor in understanding an element's reactivity and chemical properties, and is usually measured in units of kJ/mol or eV through experimental methods. The ionization energy of an element is influenced by its atomic structure, including the number of protons, distance between the nucleus and outermost electron, and electron shielding effects. The ionization energy of Lithium specifically is 520 kJ/mol, indicating the amount of energy required to remove one mole of electrons from one mole of Lithium atoms in their gaseous state.
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.

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.

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!

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.

## What is ionization energy?

Ionization energy is the amount of energy required to remove an electron from an atom or molecule in its gaseous state.

## Why is determining ionization energy important?

Determining ionization energy can help us understand the reactivity and chemical properties of an element. It also helps in predicting how an element will interact with other elements.

## How is ionization energy measured?

Ionization energy is measured in units of kilojoules per mole (kJ/mol) or electron volts (eV). It is usually measured experimentally by bombarding atoms with high energy particles and measuring the energy required to remove an electron.

## What factors affect the ionization energy of an element?

The ionization energy of an element depends on its atomic structure, specifically the number of protons, the distance between the nucleus and the outermost electron, and the electron's shielding effects from other electrons in the atom.

## What is the ionization energy of Lithium?

The ionization energy of Lithium is 520 kJ/mol. This means that it takes 520 kJ of energy to remove one mole of electrons from one mole of Lithium atoms in its gaseous state.

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