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[DESPERATE!] Frank-Hertz/Compton/De Broglie

  1. Dec 18, 2007 #1
    If you can help me on even one of these questions, it's greatly appreciated!!

    1. The problem statement, all variables and given/known data

    Use the following info to answer next 4 questions:

    The vapor of Element X produces an emission spectrum that consists of wavelengths of 400 nm, 550 nm, and 680 nm.

    1) If an unexcited sample of element x is bombarded w/ elections that have a kinetic energy of 2.00 eV, then determine the amnt of kinetic energy the e- wouild have as the exit the sample.

    2) Explain how this observation verifies Bohr's model of the atom.

    3) Determine the momentum of an emitted 570 nm light wave.

    4) Decribe what would happen to the direction and wavelength of a 570 nm light wave after it collides with:
    a) the nucleus of a large atom
    b) an electron

    ------- new Q -------

    5) Calculate the wavelength of an e- in the 4th energy level of a hydrogen atom.

    2. Relevant equations

    See below.
    And also E2 = E1 / n2
    E1 = -2.18 x 10^-18 OR -13.6 eV

    E = hc / (wavelength)
    h= 6.63 x 10^-34
    There may be more; I'm not sure which to use! Please help!

    3. The attempt at a solution

    1) This is one I have no idea... I've tried different things but keep getting the wrong answer, I believe.
    I tried finding E2, and I assumed the energy level # is 2 because it's in the visible light spectrum.
    So I used E2 = E1/n2, and then I subtracted E1 from E2, which gives me the answer to how much energy the atom can absorb. Then I tried to subtract that answer from the enegry given to me in the question, but I'm in doubt that this is correct.

    2) Bohr states that an atom is a nucleus surrounded by e- moving in circular orbits and that energy levels are quantized. This expt (Frank-Hertz) proves that e- truly receive energy in discrete amounts, and energy need is based on energy levels.

    3) p = h / (wavelength)
    p = (6.63 x 10^-34) / (570 x 10^-9)
    FINAL ANSWER: p = 1.16 x 10^-27 kg(m/s)

    4 a) Bounce back?? What happens to the wavelength? I'm stumped.
    4 b) I have no idea.

    5) En = E1/n^2
    En = -2.18 x 10^-18 / 4^2 = -1.3325 x 10^-19
    E = hc / (wavelength)
    (wavelength) = hc / E = ((6.63 x 10^-34)(3x10^8)) / (-1.3625 x 10^-19)
    FINAL ANSWER: -1.46 x 10^-6
     
    Last edited: Dec 18, 2007
  2. jcsd
  3. Dec 19, 2007 #2

    Hootenanny

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    Try converting each of the wavelengths into its respective energy.
    Correct.
    What does Compton Scattering tell you? (look at the equation).
    Correct! Assuming that you punched the numbers into your calculator correctly.
     
  4. Dec 19, 2007 #3
    Thank you so much!

    I believe that I have figured out number one.
    I found the energy of each wavelength to be:
    4.97 x 10^-19 J
    3.62 x 10^-19 J
    3.49 x 10^-19 J
    and 2.93 x 10^-19 J

    Then I found the energy of the electron (2.00 eV multiplyed by 1.6x10^-19, which equals 3.2x10^-19).

    Then I subtracted the energy of the wavelength which came closest to 3.2x10^-19 without going over, cuz I assumed maximum energy is absorbed.

    So I did: (3.2x10^-19)-(2.93x10-19) giving me a final answer of: 2.75 x 10^-20 J

    Does that last part seem correct?
    Also, is my explanation of why it verifies Bohr's model correct?

    and..

    All I have in my notes is a small section of Rutherford scattering expt, which just says that it was discovered alpha particles bounce off the nucleus, and then my notes have a small bit about Maxwell's suggestion.

    I'm trying to search google about Compton scattering, but I'm not really understanding it as we haven't learned any of those equations.

    I'm in Physics 30, by the way.

    Any more hints would be great! :) Thanks.
     
    Last edited: Dec 19, 2007
  5. Dec 20, 2007 #4

    Hootenanny

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    Looks good to me :approve:
    Sounds okay to me, but it may be improved by explicitly mentioning that the discrete wavelengths corresponding to discrete energy absorption.
    Physics 30, from what I can gather this would be similar to UK A-Levels, which would make you roughly 17? If that is the case then I wouldn't worry about Compton Scattering. To answer the final question, compare the energies of the photon to and energy of the nucleus (assume that the nucleus is at rest) and think about what would happen classically. Then do the same for the electron.
     
  6. Dec 20, 2007 #5
    Awesome, thanks! Yup, I'm 17 exactly.
    I'll try comparing the energies and see what I can do.
     
  7. Dec 20, 2007 #6
    The only part I could find energy for was the photon, by using E = hc/(wavelength)
    The nucleus has no kinetic energy, how would I find it's potential? Or does it have none?
    I am not sure how to find the energy of the electron either, since I only know the charge to be 1.6 x 10^-19. Is that equivalent to its energy?

    Here are the answers I have come up with:

    When the light wave collides with the nucleus:
    - wavelength stays the same
    - it will bounce off depending on what angle it hits at

    When the light wave collides with the electron:
    - wavelength increases (energy, therefore frequency, is lost because energy was transferred to the electron)
    - wave travels in same direction


    I had a 97% final mark in Physics 20; and a 93% so far in Physics 30 -- but this last unit is killing me! It's not making much sense at all... very frustrating. My unit exam is tomorrow - wish me luck, because I'll need it!
     
  8. Dec 20, 2007 #7

    Hootenanny

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    Using Einstein's energy-mass relationship the total energy of, say a carbon-12 nucleus, at rest is;

    [tex]E_n \approx 12\cdot(931 MeV) \approx 11 GeV[/tex]

    Now, if we calculate the associated energy of 570nm a photon;

    [tex]E_p = \frac{hc}{\lambda} = \frac{1240}{570} = 2eV[/tex]

    So we see that the energy of the nucleus is much greater (~1010) than the energy of the photon. So we have a classical analogy of a ping-pong ball (photon) colliding with a stationary bowling ball (nucleus). And as you correctly say;
    This is almost correct, but the photon won't generally continue in the same direction. The rest mass of the electron is around 0.511MeV, which is much closer to the energy of a photon.
    Good luck!! Make sure you come back and let us know how you get on!
     
  9. Dec 21, 2007 #8
    I had my exam this morning, and it went quite well!! I studied for 5 hours last night, and thanks to your help too, I find the unit was easier.
    My diploma (final) exam is coming up in a few weeks after Christmas break; so I may pop in if I have any more questions from previous units as I study!

    Thanks again :)
     
  10. Jan 23, 2008 #9
    Compton Scattering of X-Rays

    Hi all. I'm new here. Currently I'm enrolled into first year of physics science undergraduate in a local university in M'sia. I'm currently having some difficulties in my lab report concerning the above title. I've come up with the graph and i was confused. Anyone can help me with this dilemma here please...

    The task requires me to determine the change of wavelength using an x-ray unit. The graph i plotted seems very weird. How should be the graph look like. Too bad its a new practical procedure in my course and no seniors had done before.
     
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