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Question about c = vλ

  1. Dec 5, 2009 #1
    Okay, I just need to grasp what the equation C = Vλ means
    .
    Correct me if I am wrong.
    c = speed of light
    v = speed of frequency
    λ = wavelength

    So if I wanted to find the speed of the frequency of a photon with a wavelength of 450nm

    So I would use
    v = c/λ
    v = 299,792,458/4.5 * 10^(-11)
    = 6.662054622 * 10^18

    Then, if I wanted to find the energy of the photon, I would just multiply that number by planks constant (6.26 * 10^-34) and I get the energy of a quantum leap from an atom and the photon's energy itself, right?
     
    Last edited: Dec 5, 2009
  2. jcsd
  3. Dec 5, 2009 #2

    Integral

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    You seem to have a couple of problems. First to find the units on a quantity do dimensional analysis. This means to do algebra on the units of the quantities involved. In this case we have

    [tex] \nu = \frac c {\lambda} [/tex]

    the units of c are m/s the units of [itex] \lambda [/itex] are m so you have

    [tex]\nu = \frac {\frac m s} m [/tex]

    so the units for [itex] \nu [/itex] are s -1 more commonly called Hertz or Hz.

    Now for your numerical value, do you understand the meaning of the exponential term? Your number is off by many orders of magnitude. I get something like 7 e14 or 700,000,000,000,000Hz
     
  4. Dec 5, 2009 #3

    Matterwave

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    450nm is 4.5*10^-7 not -11

    Nano is 10^-9

    Also, we usually just call Frequency "frequency" instead of "speed of frequency" since there's no real speed (in the physics sense) involved.
     
  5. Dec 7, 2009 #4

    Ich

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    Integral already mentioned it, but explicitly: it's not v, it's [tex]\nu[/tex]. While v normally denotes a speed, [tex]\nu[/tex] stands for a frequency, which is quite different.
     
  6. Dec 7, 2009 #5

    atyy

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    Even more useless info:
    http://en.wikipedia.org/wiki/Nu_(letter)
     
  7. Dec 7, 2009 #6

    ideasrule

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    To avoid confusion, I'd use "f".
     
  8. Dec 7, 2009 #7

    In additon to what others mentioned...

    Since you know that [itex]E=h\nu[/itex], you could just do this to find the energy of a photon:

    [tex]\nu = \dfrac{c}{\lambda}[/tex]

    [tex]E = \dfrac{hc}{\lambda}[/tex]

    A helpful number to remember is the value of hc. So if you express your wavelength in nanometers, you can get the energy in electron volts like this:

    [tex]E = \dfrac{1240eV\times nm}{\lambda}[/tex]
     
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