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Posted by on Sunday, November 4, 2007 at 11:22am.

What is the wavelength of the transition from n=4 to n=3 for Li2+? In what region of the spectrum does this emission occur? Li2+ is a hydrogen-like ion. Such an ion has a nucleus of charge +Ze and a single electron outside this nucleus. The energy levels of the ion are -Z^2RH/n^, where Z is the atomic number

okay I saw the rydberg formula but how am i or what numbers am i supposed to plug where

  • Chemistry - , Sunday, November 4, 2007 at 11:48am

    The energy levels are
    En = -Z^2RH/n^2
    You left out the 2.
    In your case, calculate En for Z = 3, with n = 4 and again for Z = 3 and n=3. Call those energy numbers E4 and E3.
    E4 - E3 is the wave number (1/wavelength) of the photon emitted.
    Invert that to get the wavelength

  • Chemistry - , Sunday, November 4, 2007 at 11:55am

    DrWLS showed you what to do.
    1/lambda = RHZ2*[(1/N12)-(1/N22].

    RH = 1.0967758341 x 10^7
    Z = 3
    N1 = 3
    N2 = 4
    Solve for lambda.

  • Chemistry - , Sunday, November 4, 2007 at 12:22pm

    okay so it should look like this

    1/lambda= 1.0967758341 x 10^7(3^2)*[(1/3^2)-(1/4^2)]

    1.0967758341 x 10^7(9)(.0486)

    i got 4798394.274

  • Chemistry - , Sunday, November 4, 2007 at 12:59pm

    I don't get that. Your work LOOKS ok to me except your answer is 1/lambda. You need to take the reciprocal, right?

  • Chemistry - , Sunday, November 4, 2007 at 1:16pm

    so i got 1/4797297.498

  • Chemistry - , Sunday, November 4, 2007 at 1:25pm

    Plug that into your calculator; i.e., divide 1 by that large number and the answer will be the wavelength. That's what you are trying to find.

  • Chemistry - , Sunday, November 4, 2007 at 1:27pm

    One note.
    Since you are carrying all the other numbers to such high precision, I would suggest you carry the 0.0486 you obtained to the same number of significant figures.

  • Chemistry - , Sunday, November 4, 2007 at 2:07pm

    so I did 1 divided by 4797297.498

    and got 2.085 X 10^-7

  • Chemistry - , Sunday, November 4, 2007 at 2:17pm

    That's what I have. Usually the answer is expressed either in Angstrom units or in nanometers. Your answer is 208.5 nm or 2085 Angstroms. Again, if you want to carry that many places for RH, then you are justified in more placed in the answer. I don't know how your professor wants it done. Many use 1.097 x 10^7 for RH.

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