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Best_Mathematician

Photons help

  • 11 months ago
  • 11 months ago

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

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

    • 11 months ago
  3. Best_Mathematician
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    how?

    • 11 months ago
  4. .Sam.
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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

    • 11 months ago
  5. Best_Mathematician
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    so is it 6.62*10^-34/3.2? hey what is c then?

    • 11 months ago
  6. .Sam.
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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, \[\lambda=\frac{6.63 \times 10^{-3}(3 \times 10^8)}{3.2e}\]

    • 11 months ago
  7. .Sam.
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    the charge, e, is \(1.6 \times 10^{-19}C\)

    • 11 months ago
  8. Best_Mathematician
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    so is the answer 1.24*10^-6

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

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

    • 11 months ago
  11. .Sam.
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    That's the wavelength :)

    • 11 months ago
  12. Best_Mathematician
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    and the unit of wavelength is Hz right

    • 11 months ago
  13. .Sam.
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    Hertz is for frequency, Wavelength is meter,

    • 11 months ago
  14. Best_Mathematician
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    3.89∗10^(−7) meters ???

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

    • 11 months ago
  16. Best_Mathematician
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    oh ya thx...what abt 2nd one

    • 11 months ago
  17. .Sam.
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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

    • 11 months ago
  18. .Sam.
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    So the frequency plays a role here on the emission of photons.

    • 11 months ago
  19. .Sam.
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    I gotta go now, I'll leave it here, think about that :)

    • 11 months ago
  20. amistre64
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    this is beyond my learning curve :) i can recall levers and rotations and such

    • 11 months ago
  21. Best_Mathematician
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    haha thx tho

    • 11 months ago
  22. abb0t
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    I would need to refresh my memory in modern physics. So no. Lol

    • 11 months ago
  23. Best_Mathematician
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    haha ok.

    • 11 months ago
  24. abb0t
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    @Frostbite might know.

    • 11 months ago
  25. Best_Mathematician
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    well looks like audience is here...but the players are not. haha

    • 11 months ago
  26. Eyad
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    I will make some researches and see if i can help :3

    • 11 months ago
  27. abb0t
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    can't you use kirkchoffs laws here?

    • 11 months ago
  28. Jemurray3
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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.

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

    • 11 months ago
  30. Best_Mathematician
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    i do think we r using planck's equation and stuff here

    • 11 months ago
  31. Jemurray3
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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.

    • 11 months ago
  32. Jemurray3
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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 \[ \lambda = \frac{hc}{E_2-E_0} \]

    • 11 months ago
  33. Best_Mathematician
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    is this first answer?

    • 11 months ago
  34. Jemurray3
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    Yes.

    • 11 months ago
  35. Best_Mathematician
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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

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

    • 11 months ago
  37. Best_Mathematician
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    oh gotcha

    • 11 months ago
  38. Jemurray3
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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?

    • 11 months ago
  39. Best_Mathematician
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    so energy photons absorbed are all of them?

    • 11 months ago
  40. Jemurray3
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    Photons could only be absorbed if that have E = (E2-E0), E = (E1-E0), or E= (E2-E1).

    • 11 months ago
  41. Best_Mathematician
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    ohk...so thats the second answer...what abt third one

    • 11 months ago
  42. Jemurray3
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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.

    • 11 months ago
  43. Best_Mathematician
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    oh so in last two answers we dont need any calculations and stuff?

    • 11 months ago
  44. Jemurray3
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    No.

    • 11 months ago
  45. Best_Mathematician
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    what abt work function...is it the same thing as absorbing and emitting photons

    • 11 months ago
  46. Jemurray3
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    No. The work function is not related to your question at all. It is a separate effect.

    • 11 months ago
  47. Best_Mathematician
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    nice thanks...

    • 11 months ago
  48. Best_Mathematician
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    wish to give u 3 medals but just have 1

    • 11 months ago
  49. Jemurray3
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    Quite alright, medals are not important.

    • 11 months ago
  50. Best_Mathematician
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    exactly thx for explaining stuff

    • 11 months ago
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