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Which photon has the higher frequency?

  1. Apr 2, 2005 #1
    Chemistry is not my strong subject...

    A hydrogen atom undergoes a transition from the state n=3 to n=2 and then another transition from n=2 to n=1. Two photons are created due to these processes. Which photon has the higher frequency?

    AND What is meant by 'n = #' ?
     
  2. jcsd
  3. Apr 2, 2005 #2

    SpaceTiger

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    Check out this thread, keeping in mind that [tex]E=h\nu[/tex]. The principal quantum number, "n", is a property of a quantum mechanical state with certain energy. States with different "n" have different energies.
     
  4. Apr 6, 2005 #3
    I didn't understand the discussion on that site very well. The course I'm in is a physics course and they threw in some chemistry, however, we don't have a textbook, so my resource is mainly the internet...

    States with different 'n' values have different energies, but do the energies from n = 1, 2, 3 differ by a constant value [i.e. 1, 3, 5] or not?
     
  5. Apr 6, 2005 #4

    SpaceTiger

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    Actually, the energy of a state is proportional to the square of the inverse of the principal quantum number, "n". So, if we choose units in which the energy of the n=1 state is -1 (the total energy of a bound state is always negative), then the n=2,3, and 4 states will have energies of -1/4, -1/9, and -1/16, respectively. An energy greater than 0 will mean that the electron is no longer bound to the nucleus.
     
  6. Apr 6, 2005 #5
    ok got that...so then obviously an electron moving from energy state -1/9 to -1/4, it would give off more energy than one moving from -1/4 to -1, correct?
     
  7. Apr 6, 2005 #6

    SpaceTiger

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    You might want to try that subtraction again. :wink:
     
  8. Apr 6, 2005 #7
    Well I can see what I did wrong but doesn't an electron moving to a state 'closer' to the nucleus give off energy?
     
  9. Apr 6, 2005 #8

    SpaceTiger

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    If by "closer" you mean more tightly bound to the nucleus, then yes. The most tightly bound state is n=1.
     
  10. Apr 6, 2005 #9
    Yes, what I mean is that the electrons are more tightly bound to the nucleus. So, how should I do my subtraction? I should do -1/4 - -1/9 ? <-- Negative number would be required for energy released... ?
     
    Last edited: Apr 6, 2005
  11. Apr 6, 2005 #10

    SpaceTiger

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    When it moves from state n=3 to n=2, does the electron's energy increase or decrease (based on what I already said)? If energy is conserved, what does this mean for the photon that's emitted?
     
  12. Apr 6, 2005 #11
    When moving from state n = 3 to n = 2, the electron's energy decreases... it loses some of the energy to the escaping photon...that's what I'm understanding..
     
  13. Apr 6, 2005 #12

    SpaceTiger

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    That's right. So, to conserve energy, the atom must emit a photon of light!
     
  14. Apr 6, 2005 #13
    Exactly, soo, but then isn't my subtraction correct?

    The only thing I'm seeing wrong with my subtraction is that for 'energy' I should have a negative answer...
     
  15. Apr 6, 2005 #14

    SpaceTiger

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    In our units, the energy change from n=3 to n=2 is (-1/9)-(-1/4)=5/36. This is the energy that's released as light. It's the change in energy of the electron that's negative: (-1/4)-(-1/9)=-5/36.
     
  16. Apr 6, 2005 #15
    Ahh, I see what you mean. E2 - E1 = energy released as light, while E1 - E2 = change in energy of the electron.
     
  17. Apr 6, 2005 #16
    That would mean that more energy is released from n = 2 to n = 1, than is from n = 3 to n = 2, would it not?
     
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