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anonymous

  • one year ago

Please help I will medal! I am confused on how to solve this question: Using the conversion factor from eV (electron volts) to joules, determine which energy line for an electron dropping from one energy level to another would show a line at 435 nm in an emission spectrum. Use the formulas and the energy level information below to answer the question. Needed constants: 1.00eV=1.6*10^(-19) J C= 3.0*10^(8) M/J H= 6.63*10^(-34) J*S Energy Level Values: E6:E= -0.378eV E5:E= -0.544eV E4:E= -0.850eV E3:E= -1.51eV E2:E= -3.403V

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  1. anonymous
    • one year ago
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    @glutich @abdul_shabeer @Kitten_is_back @adajiamcneal @stonewoods @HazelLuv99 @Callisto

  2. anonymous
    • one year ago
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    @gamer56 @KIT-KAT-KATE @Anon101 @JFraser

  3. Gamer56
    • one year ago
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    Hello Ella, First do you know any of the steps to completing this problem ?

  4. anonymous
    • one year ago
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    No, this question is unlike any I have seen before.

  5. Gamer56
    • one year ago
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    Ok Ella, I will help you. Please give me a second.

  6. anonymous
    • one year ago
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    Thank you so much!

  7. Gamer56
    • one year ago
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    What you do is Use E=hf f=E/h

  8. JFraser
    • one year ago
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    the energy of the emitted wave must match the energy \(difference\) between the two levels that it falls between

  9. Gamer56
    • one year ago
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    E=difference of 2 energy levels h=plank's constant f=frequency of emission

  10. Gamer56
    • one year ago
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    Is this making any sense so far ?

  11. JFraser
    • one year ago
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    First, you need to find the energy of the 435nm wave

  12. anonymous
    • one year ago
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    Yes it is. Alright.

  13. anonymous
    • one year ago
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    Sorry the yes it is was to @gamer56

  14. Gamer56
    • one year ago
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    @Ella31224 Ok. Im sorry ella, But im having internet problems. Maybe @JFraser Can finish helping you ?

  15. anonymous
    • one year ago
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    @JFraser If you could help me I would really appreciate it.

  16. JFraser
    • one year ago
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    Can you find the energy (E) of the 435nm wave?

  17. anonymous
    • one year ago
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    Would the equation be E=(6.63*10^(-34) J*S)(435)?

  18. JFraser
    • one year ago
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    almost, you're forgetting that \(1nm = 1*10^{-9}m\), so you need to include the exponent

  19. JFraser
    • one year ago
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    \(435nm = 435*10^{-9}m\)

  20. JFraser
    • one year ago
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    wait, stop that

  21. anonymous
    • one year ago
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    Stop what?

  22. JFraser
    • one year ago
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    I didn't see you'd actually put up the wrong equation

  23. anonymous
    • one year ago
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    What equation should I use?

  24. JFraser
    • one year ago
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    it should be \(c = \lambda * \nu\), where c is the speed of light and \(\lambda\) is the wavelength

  25. JFraser
    • one year ago
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    that will get you the \(frequency\), which is \(\nu\). That frequency gets plugged into \(E = h*\nu\)

  26. anonymous
    • one year ago
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    Alright so V= (3.0*10^(8) M/J)/(435 x 10^-9 m)?

  27. JFraser
    • one year ago
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    the units are m/s and m, but yes

  28. JFraser
    • one year ago
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    that will get you the frequency. Plug that frequency into the energy equation and get the energy of the wave

  29. JFraser
    • one year ago
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    that energy will be in \(joules\), so you need to use the eV-J conversion to find the energy in eV

  30. anonymous
    • one year ago
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    Ok can you solve it with me just ensure I get the right answer?

  31. JFraser
    • one year ago
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    What did you get, I'll check

  32. JFraser
    • one year ago
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    the frequency should be a huge number, like \(10^{14}\)

  33. anonymous
    • one year ago
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    yeah Im not getting a very large number..

  34. JFraser
    • one year ago
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    if you're not careful, the order of operations on the calculator will give you a really random answer

  35. anonymous
    • one year ago
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    Can you show me your order of operations because I am not getting a large number.

  36. anonymous
    • one year ago
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    I got 0.6896551724137931 as my frequency

  37. anonymous
    • one year ago
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    @gamer56

  38. anonymous
    • one year ago
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    @Smita12 Could you help me?

  39. anonymous
    • one year ago
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    :(

  40. JFraser
    • one year ago
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    you need to make sure that the exponents get handled the right way. The conversion should look like\[\frac{3*10^8\frac{m}{s}}{435*10^{-9}m}\]

  41. anonymous
    • one year ago
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    Was that the incorrect frequency?

  42. JFraser
    • one year ago
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    that will \(get \space you\) the frequency, but you have to do out the math

  43. anonymous
    • one year ago
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    Again I got 6.8965517e+14..

  44. JFraser
    • one year ago
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    right "e+14" is the calculator's way of telling you scientific notation

  45. anonymous
    • one year ago
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    so it is 6^14?

  46. JFraser
    • one year ago
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    so you did get the right frequency

  47. JFraser
    • one year ago
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    there's a big difference between \(6^{14}\), and \(6*10^{14}\)

  48. anonymous
    • one year ago
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    Sorry 6 x 10^14

  49. JFraser
    • one year ago
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    that's the frequency

  50. anonymous
    • one year ago
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    Awesome! So now plug it into the energy equation?

  51. anonymous
    • one year ago
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    E = (6.63 x 10^(-34) J*S)(6 x 10^14) correct?

  52. JFraser
    • one year ago
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    correct

  53. JFraser
    • one year ago
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    and you should get an energy that's pretty small

  54. anonymous
    • one year ago
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    3 x 10^-19?

  55. JFraser
    • one year ago
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    sounds about right

  56. JFraser
    • one year ago
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    that's in \(joules\), but the energy levels you're given are in eV. so use the conversion to find the energy of that wave in eV

  57. anonymous
    • one year ago
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    Alright

  58. anonymous
    • one year ago
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    The closest eV to my answer is E3:E= -1.51eV.

  59. anonymous
    • one year ago
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    What answer did you get?

  60. JFraser
    • one year ago
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    it's not the energy level that's important, because the wave is created by the electron moving \(between\) two levels. You need to find two levels that have a \(difference\) of 1.5eV

  61. anonymous
    • one year ago
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    Alright so how would I go about doing this?

  62. JFraser
    • one year ago
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    when you calculated the frequency, you cut off a few too many digits. the frequency should be closer to 6.8*10^14Hz, which makes the energy closer to 4.6*10^-19J

  63. JFraser
    • one year ago
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    that makes the eV difference about 2.85eV

  64. anonymous
    • one year ago
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    Alright and is that the correct answer?

  65. JFraser
    • one year ago
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    again, it's the \(difference\) between two levels that must equal about 2.85 eV

  66. JFraser
    • one year ago
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    the energy levels sort of look like this|dw:1432832938799:dw|

  67. JFraser
    • one year ago
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    |dw:1432833072141:dw| when the wave is created, the energy \(difference\) between two levels is equal to the energy of the wave. You know the energy of the wave is about 2.85eV, so find two levels separated by about 2.85eV

  68. anonymous
    • one year ago
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    So the two levels would be E3 and E2?

  69. JFraser
    • one year ago
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    3.85eV - 1.5eV is only about 2.3eV, can there be a better choice?

  70. anonymous
    • one year ago
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    E4 and E2?

  71. JFraser
    • one year ago
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    that's only 2.6 (I like E5 - E2)

  72. anonymous
    • one year ago
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    Awesome thank you! So the answer to my question is E5 - E2?

  73. JFraser
    • one year ago
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    that's what it seems to me

  74. anonymous
    • one year ago
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    Thank you so much! You helped me learn a lot!

  75. JFraser
    • one year ago
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    YW

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