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Best_Mathematician Group Title

Photons help

  • one year ago
  • one year ago

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  1. aaronq Group Title
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    use plancks equation: E=hc/lambda

    • one year ago
  2. rohankv Group Title
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    Well use plank's equations.....however answer depends on geometry of molecule in case of part B & part C

    • one year ago
  3. Best_Mathematician Group Title
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    how?

    • one year ago
  4. .Sam. Group Title
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    They have given you the energy of \(E_2\), so by using equation \(E=hf\) and \(c=f \lambda\), we can form \[E_2=\frac{hc}{\lambda} \\ \\ \lambda=\frac{hc}{E_2}\] h=plank's constant c=speed of light \(E_2\)=energy Use that to find the wavelength at that energy

    • one year ago
  5. Best_Mathematician Group Title
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    so is it 6.62*10^-34/3.2? hey what is c then?

    • one year ago
  6. .Sam. Group Title
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    "c" is the speed of light, which has a value of \(3 \times 10 ^8m/s\) and keep in mind that, E is not just 3.2, because its eV, you have to multiply by the charge too, \[\lambda=\frac{6.63 \times 10^{-3}(3 \times 10^8)}{3.2e}\]

    • one year ago
  7. .Sam. Group Title
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    the charge, e, is \(1.6 \times 10^{-19}C\)

    • one year ago
  8. Best_Mathematician Group Title
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    so is the answer 1.24*10^-6

    • one year ago
  9. .Sam. Group Title
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    How do you get that? should be \[\lambda=\frac{6.63 \times 10^{-34}(3 \times 10^8)}{3.2(1.6 \times 10^{-19})}\]

    • one year ago
  10. Best_Mathematician Group Title
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    ya and then i solved it using calculator...i got\[3.89 * 10^{-7}\]

    • one year ago
  11. .Sam. Group Title
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    That's the wavelength :)

    • one year ago
  12. Best_Mathematician Group Title
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    and the unit of wavelength is Hz right

    • one year ago
  13. .Sam. Group Title
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    Hertz is for frequency, Wavelength is meter,

    • one year ago
  14. Best_Mathematician Group Title
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    3.89∗10^(−7) meters ???

    • one year ago
  15. .Sam. Group Title
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    yes if you want to convert that into nanometers, then its \[(3.89 \times 10^{-7}) \times 10^9 = 389nm\]

    • one year ago
  16. Best_Mathematician Group Title
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    oh ya thx...what abt 2nd one

    • one year ago
  17. .Sam. Group Title
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    The second one is referring to work function energy \(\phi\), work function is the minimum energy needed to remove an electron from a solid to a point immediately outside the solid surface. We express that by \[\phi=hf_o\] \(f_o\)= minimum frequency to cause a photoelectric emission

    • one year ago
  18. .Sam. Group Title
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    So the frequency plays a role here on the emission of photons.

    • one year ago
  19. .Sam. Group Title
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    I gotta go now, I'll leave it here, think about that :)

    • one year ago
  20. amistre64 Group Title
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    this is beyond my learning curve :) i can recall levers and rotations and such

    • one year ago
  21. Best_Mathematician Group Title
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    haha thx tho

    • one year ago
  22. abb0t Group Title
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    I would need to refresh my memory in modern physics. So no. Lol

    • one year ago
  23. Best_Mathematician Group Title
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    haha ok.

    • one year ago
  24. abb0t Group Title
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    @Frostbite might know.

    • one year ago
  25. Best_Mathematician Group Title
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    well looks like audience is here...but the players are not. haha

    • one year ago
  26. Eyad Group Title
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    I will make some researches and see if i can help :3

    • one year ago
  27. abb0t Group Title
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    can't you use kirkchoffs laws here?

    • one year ago
  28. Jemurray3 Group Title
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    A couple of things... first, if you just plug in the energy E2, you're going to get the wrong answer. You instead need to use the energy difference E2-E0.

    • one year ago
  29. Best_Mathematician Group Title
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    i really dont know how to do this...so if anyone can ex[lain it will be great

    • one year ago
  30. Best_Mathematician Group Title
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    i do think we r using planck's equation and stuff here

    • one year ago
  31. Jemurray3 Group Title
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    Okay here's way it works. This molecule has three energy levels that you're considering -- E0, E1, E2. These are the only energies it can have -- nothing in between. Photons are just traveling packets of energy. Therefore, if the atom is in the ground state E0, it might absorb a passing photon and be promoted up to a higher energy level. This will ONLY work, though, if the photon is carrying exactly as much energy as is required. If the photon has too much energy, it won't work. If it has too little, it won't work either. The amount of energy required to promote the atom from state E0 to state E2 is E2-E0.

    • one year ago
  32. Jemurray3 Group Title
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    For a photon, \[ E_{photon} = \frac{hc}{\lambda} \] or \[ \lambda = \frac{hc}{E_{photon}} \] This tells you that if you want to promote the atom from state E0 to state E2, you need a photon of wavelength \[ \lambda = \frac{hc}{E_2-E_0} \]

    • one year ago
  33. Best_Mathematician Group Title
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    is this first answer?

    • one year ago
  34. Jemurray3 Group Title
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    Yes.

    • one year ago
  35. Best_Mathematician Group Title
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    so the guy before who explained me the answer is wrong? we would need to include E2-E0 to get wavelength

    • one year ago
  36. Jemurray3 Group Title
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    Yes. His equations weren't wrong but you need the energy difference, not just the energy of the excited state.

    • one year ago
  37. Best_Mathematician Group Title
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    oh gotcha

    • one year ago
  38. Jemurray3 Group Title
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    For the second part, because the only allowed energies are E0, E1, and E2, the only possible promotions are from E0->E1, E0->E2, and E1->E2. Each transition has a particular energy difference. Therefore, each transition can be caused only by a photon of a precise energy. It is these photons that can be absorbed by your atom. Do you understand?

    • one year ago
  39. Best_Mathematician Group Title
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    so energy photons absorbed are all of them?

    • one year ago
  40. Jemurray3 Group Title
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    Photons could only be absorbed if that have E = (E2-E0), E = (E1-E0), or E= (E2-E1).

    • one year ago
  41. Best_Mathematician Group Title
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    ohk...so thats the second answer...what abt third one

    • one year ago
  42. Jemurray3 Group Title
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    The third is the reverse process. If the atom is in an excited state, it can only transition down in precise ways. That is, the only allowed downward transitions are E2->E1, E2->E0, and E1->E0. When an atom undergoes a downward transition, it emits a photon whose energy is equal to the energy difference of the transition.

    • one year ago
  43. Best_Mathematician Group Title
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    oh so in last two answers we dont need any calculations and stuff?

    • one year ago
  44. Jemurray3 Group Title
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    No.

    • one year ago
  45. Best_Mathematician Group Title
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    what abt work function...is it the same thing as absorbing and emitting photons

    • one year ago
  46. Jemurray3 Group Title
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    No. The work function is not related to your question at all. It is a separate effect.

    • one year ago
  47. Best_Mathematician Group Title
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    nice thanks...

    • one year ago
  48. Best_Mathematician Group Title
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    wish to give u 3 medals but just have 1

    • one year ago
  49. Jemurray3 Group Title
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    Quite alright, medals are not important.

    • one year ago
  50. Best_Mathematician Group Title
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    exactly thx for explaining stuff

    • one year ago
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