Chemistry

posted by .

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 -

    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 -

    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 -

    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 -

    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 -

    so i got 1/4797297.498

  • Chemistry -

    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 -

    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 -

    so I did 1 divided by 4797297.498

    and got 2.085 X 10^-7

  • Chemistry -

    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.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Chemistry

    What is the wavelength of the transition from n=4 to n=3 for Li2+?
  2. Chem

    What is the wavelength of the transition from n=4 to n=3 for Li2+?
  3. chemistry

    What is the wavelength of the transition from n = 2 to n = 1 for Li2+?
  4. chemistry

    Hydrogen-like ions are those with only one electron such as He+ and Li2+. The Balmer formula for the transitions in these ions can be written as v=Z^2 cR(1/(n_1^2 )-1/(n_2^2 )). Use this formula to calculate the lowest-energy transition …
  5. Chemistry

    Considering the emission of the hydrogen atom, a. what is the wavelength of the light emitted by the transition from n = 2 to n = 1. b. In what region of the electromagnetic spectrum does this radiation belong?
  6. Chemistry

    Atoms of ionized lithium gas (Li2+) are struck by neutrons moving at a velocity of 1.46e5 m/s. Calculate the shortest wavelength in the emission spectrum of Li2+ under these circumstances. You can assume that all electrons start in …
  7. chemistry

    Atoms of ionized lithium gas (Li2+) are struck by neutrons moving at a velocity of 1.46e5 m/s. Calculate the shortest wavelength in the emission spectrum of Li2+ under these circumstances. You can assume that all electrons start in …
  8. chemistry

    A sample contains Hydrogen atom ,He+ ion, Li2+, & Be3+ ion. In H atom electron is present in 8th orbit, in He2+ e- is present in 6th orbit, Li2+ in 5th orbit& in Be3+ electron is present in 4th orbit. All the atoms are de-excited to …
  9. Chemistry

    Consider the Li2+ ion. (a) Calculate the wavelength of light emitted when the electron jumps from the n=4 level to the n=2 level. (b) What is the ionization energy of Li2+ ?
  10. Chemistry

    The Li2+ ion is hydrogenic and has a Lyman series at 740747 cm-1, 877924 cm-1, 925933 cm-1,and beyond. Show that the energy levels are of the form –hcR/n^2 and find the value of R for this ion. Go on and predict the wavenumbers of …

More Similar Questions