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Kinetic Enegry and Hydrogen atoms

  1. Nov 9, 2008 #1
    1. The problem statement, all variables and given/known data

    a) What is the minimum kinetic energy in electron volts that an electron must have to be able to ionize a hydrogen atom (that is, remove the electron from being bound to the proton)? I got the answer here which is 13.6 ev

    b) If electrons of energy 12.8 eV are incident on a gas of hydrogen atoms in their ground state, what are the energies of the photons that are emitted by the excited gas?
    b-1)Energy of highest-energy photon: I got the answer here which is 12.8
    b-2)Energy of next highest-energy photon:
    b-3)Energy of next highest-energy photon:
    b-4)Energy of next highest-energy photon:
    b-5)Energy of next highest-energy photon:
    b-6)Energy of lowest-energy photon:


    I Just need help with ( b-2 to b-6) Thanks
     
  2. jcsd
  3. Nov 9, 2008 #2

    Redbelly98

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    The answer depends on details of your solution to b-1. In particular, what energy level (n value) is the Hydrogen excited to by the 12.8 eV energy?
     
  4. Nov 9, 2008 #3
    am sorry but that 12.8 was a guess for b-1 and i dont know how to find that.
     
  5. Nov 9, 2008 #4
    Question (b) is a bad question: the energies of the photons emitted by an actual gas form a continuum because of the motion of the atoms, their collisions, and the fact the excited states are unstable. So you can't even talk about the energy of the "next highest-energy" photon.

    But setting that aside, what is meant here, no doubt, is that the 12.8 eV brings you to a particular excited state with a certain value of n, call it n_0 (easy to calculate). From n_0 the atom could make a transition to the ground state, giving back a 12.8 eV photon. But it could also make a transition to the n=2 state, emitting a less energetic photon. It could also "fall" to the n=3 state, emitting an even less energetic photon. And, once an atom is in the n=3 state, it could fall to the ground state from there, emitting yet another photon....etc.
     
  6. Nov 9, 2008 #5
    so is there a specific formula i should use to calculate these answers?
     
  7. Nov 9, 2008 #6
    Yes...don't you know a formula for the energies of hydrogen levels, as a function of "n"?

    The question clearly assumes you know this, so look it up in your textbook!
     
  8. Nov 9, 2008 #7
    the formula i know is -13.6/n^2 oh so instead of 13.6 i would use 12.8?
     
  9. Nov 9, 2008 #8
    No, it is and remains -13.6eV/n^2.
     
  10. Nov 9, 2008 #9
    Hint: the ground state corresponds to n=1. How much energy difference is there between n=1 and n=2?

    You want to find a n_0 such that the difference between the levels n_0 and n=1, is about 12.8 eV.
     
  11. Nov 9, 2008 #10
    so n0 - n1 = 12.8 and n1 is -13.6 so n0= 12.8-13.6 = -0.8
     
  12. Nov 9, 2008 #11
    Do you understand the equation that says the energy of level n is -13.6eV/n^2?
     
  13. Nov 9, 2008 #12
    it doesnt look like it cause the teacher we have is horrible. i barely understand the question sorry.
     
  14. Nov 9, 2008 #13
    It says the energy of level n=1 is -13.6 eV, for n=2 the energy is -13.6/4=-3.4 eV, etc.
    So the difference in energy between these two levels is (-3.4--13.6)eV=10.2 eV.

    Now find n_0 such that -13.6/n_0^2+13.6=12.8 eV.
     
  15. Nov 9, 2008 #14
    ok so n0^2= 13.6/0.8 = 17 then sqrt 17 = 4.12 which n0
     
  16. Nov 9, 2008 #15
    Yes, so presumably n=4 is meant as the level in which the hydrogen atoms end up (n can only be an integer!)
     
  17. Nov 9, 2008 #16
    ok good so for b-2 i would use n= 3 so -13.6/9+13.6 and that gave me 12.089
    for b-3 n=2 so i would get 10
    b-4 i would use n=1 so now i would get 0 which gave me a wrong answer the first 2 were right how come
     
  18. Nov 9, 2008 #17
    You can go from n=4 to n=2 also
     
  19. Nov 9, 2008 #18
    so i should get the same 3 answer again? if so its not working.
     
  20. Nov 9, 2008 #19
    No, your b-3 answer is for n=2 to n=1.
     
  21. Nov 9, 2008 #20
    ok so for b-4 i would repeate the same steps. start at n=4 so i will get 12.8 again, and for b-5 i would get 10 again and for b-6 i would get 12? is that right?
     
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