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Electron transition question

  1. Apr 10, 2009 #1
    E of nlow to nhigh= -2.178*10^-18 * Z^2/nhigh^2 - nlow^2

    find the energy needed to remove an electron completely from a hydrogen atom and the energy needed to remove one mol of electrons from one mol of hydrogen atoms

    Z= 1 nhigh= infinity nlow=1

    E = 2.178*10^-18 J
    one mole of electrons = 2.178*10^-18 * 6.022*10^23 = 1.312*10^6 J/mol
    Are these right??

    And use the above formula to find the value of Z for an ion whose 2 to 1 transition is associated with a wavelength of 13.4nm.

    lamba of 2 to 1= hc/(2.178*10^-18 * Z^2/nhigh^2 - nlow^2)
    z=sqrt(hc(nhigh^2-nlow^2)/(13.4*10^-9)(2.178*10^-18))
    Z= 4.5
    or 5
    Is this right??

    Thank you guys so much for your help.
    Stephen
     
  2. jcsd
  3. Apr 11, 2009 #2

    Borek

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    Staff: Mentor

    Try to use TEX or at least brackets, it doesn't look OK at the moment.
     
  4. Apr 11, 2009 #3

    Redbelly98

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    Homework Helper

    Hello Stephen,

    Looks good on the single hydrogen atom and mole-of-hydrogen question.

    But it looks like there's an error somewhere in the Z question. The correct answer is very close to an integer.

    I think the problem is with the expressions you are writing. There should be a term

    ( 1/nhigh^2 - 1/nlow^2 )​

    but instead you have

    nhigh^2 - nlow^2​

    ???

    Regards,

    Mark

    p.s. Borek is correct, it's better to at least use brackets (parantheses) to express things properly and avoid confusion.
     
  5. Apr 11, 2009 #4
    Z^2/nhigh^2 - nlow^2
    is the same thing as z^2*(1/nhigh^2 - 1/nlow^2)
     
  6. Apr 11, 2009 #5

    Redbelly98

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    Homework Helper

    Okay, but in your expression

    z = sqrt(...)​

    it has mysteriously become, literally,

    (nhigh^2-nlow^2)​

    and that is wrong.

    Try keeping it as

    (1/nhigh^2 - 1/nlow^2)​

    Also, you might find it easier to figure out what the energy is for 13.4 nm, and then work with the energy equation (1st equation of your 1st post).
     
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