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Atomic structure

  1. Oct 1, 2008 #1
    If a photon of wavelength 663nm is incident on a hydrogen atom in its fourth excited state(n=5),then
    A.the atom doesnt absorb the photon
    B.the atom absorbs the photon but is not ionized
    C.the atom absorbs the photon and is ionized,with the electron having zero kinetic energy after ionization
    D.the atom absorbs the photon and is ionized,with the electron having kinetic energy 1.33eV after ionization.

    the answer is D

    but i couldnt approach tat answer when im using 1/lambda=R(1/n1-1/n22)
    after sub in the values into this equation
    whole thing seems to be wrong...as 1/25-1/25=0

    how can i know tat the photon with wavelength 663nm is able to ionize the atom??
    and how to find the electron's kinetic energy after ionization???
     
  2. jcsd
  3. Oct 1, 2008 #2

    Redbelly98

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    1/λ = R (1/n12 - 1/n22)

    Start with, what wavelength would ionize the atom? (Use the above equation, thinking carefully about what n1 and n2 would be).

    How much energy does a photon of that wavelength have?

    Then, does a 663 nm photon have at least that much energy?
     
  4. Oct 1, 2008 #3
    let n2=0 and n1=25
    then i get the wavelength as 2278nm
    this is the wavelength for ionization of tat atom
    rite??
    so it means tat 663nm couldnt ionize the atom??
     
  5. Oct 1, 2008 #4

    Redbelly98

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    Actually, we let 1/n22=0 and 1/n12=25
    (n2=∞ and n1=5)

    Yes. Or more correctly, that's the threshhold wavelength for ionization of the atom. Any photons with more energy could also ionize the atom.

    You should figure out the energy in photons of 2278nm and 663nm in order to understand what is going on. Or, realize that photons have more energy if the wavelength is ___________. (longer or shorter?)

    p.s. I'm logging off soon, good luck!
     
  6. Oct 1, 2008 #5
    oo!!
    i got it !!
    thnx!!
    but how bout the kinetic energy???
     
    Last edited by a moderator: Oct 1, 2008
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