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How can i find the absorption/emission transitions from Energy levels

  1. May 20, 2014 #1
    1. The problem statement, all variables and given/known data

    Question 1)

    For absorption transitions:

    At n=1 the energy is 0.0eV (electron volts), at n=2 the energy is 3eV, at n=3=5eV

    Question 2)

    For emission transisions:

    At n=1 the energy is 0.0eV, n=2 energy is 3eV and at n=3 energy is 5eV

    Question 3)

    The atom has energy levels n=1(0eV),n=2(2eV) and n=3(5eV). A photon has energy E=3eV which corresponds to spectral line with λ=414nm. Where would this spectral line be observed? absorption spectrum, emission spectrum, both or neither?

    2. Relevant equations

    ΔE between the energy levels for an absorption transition ΔE= E1 - E2 etc..
    ΔE between the energy levels for an emission transition ΔE=E3 - E2 etc..

    λ=c*h/ΔE

    3. The attempt at a solution

    Question 1)

    So to find the difference in energy ΔE between the energy levels for an absorption transition at ΔE=E1 - E2, then ΔE= 0.0eV - 3eV = -3eV but in my notes this answer is positive 3eV what am i doing wrong?

    Question 2)

    To find emission transition at ΔE=E3 - E2 , then ΔE= 5eV-3eV= 2eV but in my lecture notes the answer given is negative -2eV i do not see how this is possible and where i went wrong?

    Question 3)

    Since E=3eV, I would assume it would be an emission spectrum because ΔE=E3 - E2=3eV but shouldn't the 3eV be negative because for an emission transion ΔE=E3 - E2 in question 2 the answer given was negative.
     
  2. jcsd
  3. May 20, 2014 #2
    You sound like the guy in the joke that was holding a pencil with the lead end up an wondering why the lead was at the wrong end of the pencil. Try flipping the pencil around.
     
  4. May 20, 2014 #3

    DrClaude

    User Avatar

    Staff: Mentor

    If an atom absorbs a photon, the energy of the photon is transferred to the atom, therefore one has ##\Delta E >0## for the atom. Conversely, when an atom emits a photon, the photon carries away some of the energy of the atom, and therefore ##\Delta E < 0##.

    If the emitted photon has an energy of +3 eV, by how much did the energy of the atom change?
     
  5. May 20, 2014 #4

    I do not understand your question properly but am guessing -3eV since you said for emissions transitions are ##\Delta E < 0## does that mean for absorption transition its always E2 -E1 so i get a positive answer and E1 - E2 so i get a negative answer for emission transition for questions 1 and 2. But then in question 3 i need to find the type of transition for E=3eV since this is positive it would an absorption transition. That can not be possible since absorption transition always begin from ground state. Confused lol
     
  6. May 20, 2014 #5
    How can i solve the issues in the OP with the three questions please any help appreciated. Thanks
     
  7. May 20, 2014 #6

    TSny

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    Homework Helper
    Gold Member

    The symbol ΔE represents the change in energy of the atom.

    So, ΔE = Efinal of atom - Einitial of atom.

    This is true for both absorption and emission.

    You just have to make sure that, for a specific transition, you know which energy level is the initial level and which is the final level.
     
  8. May 21, 2014 #7
    Thanks for the help :), makes sense now in regards to my problems with question 1 and question 2. But in question 3:

    The atom has energy levels n=1(0eV),n=2(2eV) and n=3(5eV). A photon has energy E=3eV which corresponds to spectral line with λ=414nm. Where would this spectral line be observed? absorption spectrum, emission spectrum, both or neither?

    Since it is obvious that the photon of energy E=3eV belongs to an emission spectrum why is the 3eV positive? because since ΔE = Efinal of atom - Einitial of atom in this case it would be the Final energy 2eV - Initial energy 5eV = -3eV
     
  9. May 21, 2014 #8

    DrClaude

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    Staff: Mentor

    The problem is not well formulated. Of course, if you are detecting a photon of energy 3 eV, that would correspond to emission. But a "spectral line at λ=414nm" doesn't necessarily mean that a photon of that energy has been detected.

    By convention, photon energies are taken to be positive, ##E = h \nu##. The ##\Delta E## for the atom will be negative if the photon was emitted, and positive if the photon was absorbed.
     
  10. May 22, 2014 #9
    Thanks for the help, i understand the issues i had earlier.
     
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