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Easy - determining n from orbital radius

  1. Jun 2, 2008 #1
    1. The problem statement, all variables and given/known data
    For the hydrogen atom, determine the principle quantum number corresponding to an orbital of radius Rn=0.01mm

    2. Relevant equations
    I'm using the following formula:

    Rn=(n^2)(Ao)/Z

    Where Rn is the orbital radius (0.01mm),
    n is the principle quantum number (wanted)
    Ao is a constant (0.5*10^-10)
    and Z is the atomic number (for hydrogen =1)

    3. The attempt at a solution
    By converting Rn into metres and subbing in all the other values I obtain a value of n=447.2135955.

    Obviously there are two discreptancies:
    1) n is not a whole number
    2) it is abnormally large

    Can anybody show me where I have gone wrong?
     
  2. jcsd
  3. Jun 2, 2008 #2

    Dick

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    i) it's not a whole number because 0.01mm is an approximate value. ii) It's abnormally large because an 0.01mm hydrogen atom is abnormally large. It doesn't seem out of line to me.
     
  4. Jun 2, 2008 #3
    Would I answer this with a rounded value then?

    A lead on question asks: for this value of principle quantum number, find the correspongind excited state energy.

    Thus E=-13.6Z^2/(n^2) becomes

    E=-13.6/(447.2135955^2)=-6.8*10^-5 eV

    (notice for this part I did not use the rounded value)
     
  5. Jun 2, 2008 #4

    Dick

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    There is no orbital at exactly 0.01mm. So yes, you round. It shouldn't matter much.
     
  6. Jun 3, 2008 #5
    I have another question, slightly different topic

    -Calculate the first excited state energies of He+, Li2+ and Be3+ compare them with the ground state energy of hydrogen.

    ground state means n=1, but how do i find n for the first excited state? does it differ for each of the 3 cases above?
     
    Last edited: Jun 3, 2008
  7. Jun 3, 2008 #6

    Dick

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    Yes, n=1 in all those cases. Look at your formula. What could change?
     
  8. Jun 3, 2008 #7
    atomic number is going to be different for each of those cases. But since I want to find the first excited state of He+.... don't I need to use the relevant principle quantum number?

    Not sure if I've missed something here, but how do I determine the principle quantum number (n) for the first excited state in the three above cases?
     
  9. Jun 3, 2008 #8

    Dick

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    They are all hydrogen-like one-electron atoms. The first excited state is n=2. Just like hydrogen. (Sorry if I said n=1 before, I missed the 'excited state'). The n's designate the state. n=1 is ground. The first state above ground is the first excited state. Must be n=2.
     
  10. Jun 4, 2008 #9
    cool, alot clearer now!
     
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