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Homework Help: Determining ionization energy of Lithium.

  1. Oct 30, 2017 #1
    1. The problem statement, all variables and given/known data
    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

    2. Relevant equations
    Maybe:
    EN=-(2.18E-18)(Z2/N2)

    3. 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.
     
  2. jcsd
  3. Oct 31, 2017 #2

    mfb

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    How many electrons does lithium have if two were removed already? Where are the remaining electrons?
    Quantities used in formulas are explained at the place where you found the formula. There is no need to guess.
     
  4. Oct 31, 2017 #3
    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!
     
  5. Oct 31, 2017 #4

    mfb

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    Right.
    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.
    Sure, the first and second ionization energies are lower.
    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.
     
  6. Oct 31, 2017 #5
    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!
     
  7. Oct 31, 2017 #6

    mfb

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    I'm not sure what you mean by the second expression.
    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.
     
  8. Oct 31, 2017 #7

    Borek

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    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.
     
  9. Oct 31, 2017 #8
    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.
     
  10. Nov 1, 2017 #9

    mfb

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    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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