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Number of photons emited physics 30

  1. Jan 1, 2008 #1
    [SOLVED] number of photons emited ..physics 30

    1. hi im having a few problems figuring what to do in this questions ..it says:

    A Laser Emits light of wavelenght 6.3 x 10^-7 m.


    2. i need to solve for :
    A) What is the enrgy of each photon Emitted?
    B) The total power output of the laser is 0.50 w.How many photons are emitted in a 3.0 s interval?

    3.i habe done the folowing
    A)
    to find the enrgy i use the equation
    E= hc/λ
    =(6.63 x10^-34)(3.00 x 10^8)/6.3 x 10 ^-7
    =3.16 x 10^-19J

    B) i Know the power in wats is 0.50 w,and the time is 3.0s..the frequency is
    E=hf
    F= E/h
    =3.16 x 10^-19/6.63 x 10^-34
    =5 x 10^14 Hz
    but im not sure what is the equation to find the # of photons..
    could someone help me thanks..
     
  2. jcsd
  3. Jan 1, 2008 #2

    G01

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    In order to find the number of photons you need to know how energy relates to power.

    HINT:

    You know how the total energy emitted relates to the number of photons.

    Can you use the power rating of the laser to find the total energy emitted over the 3s interval?
     
  4. Jan 1, 2008 #3
    well, energy is related to power by
    p=E/t
    E=p.t
    so it will be
    =0.50 w x 3.0s
    =1.5
    ....so this energy released will be the #of photons? im not sure how they are realted (energy with the emited photons)
     
    Last edited: Jan 1, 2008
  5. Jan 1, 2008 #4

    G01

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    That is the relationship between the energy and the number of photons.

    So, you know there are 3.16X10^-19 J of energy in one photon. How many photons do you need to get 1.5J?
     
  6. Jan 2, 2008 #5
    oh i see now i get it ..so i will need 4.74 x 10^-19 photons to get the 1.5J...

    thanks..
     
  7. Jan 2, 2008 #6

    G01

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    Anytime. :smile:
     
  8. Jan 2, 2008 #7
    Wait just a second: you have about 3*10^-19 joules per photon, but you have to take the inverse to get how many photons there are per joule, and then multiply that by the total energy output to find the number of photons emitted in the interval. There's going to be a lot of photons emitted by a laser turned on for three seconds, not a fraction of 1 photon!
     
  9. Jan 3, 2008 #8
    well i took the 3.16 x 10^-19J and times it by the 1.5J to get the the number of photons 4.74 x 10^-19photons isn't that the answer??...
    Now what you mean is to time that by the output energy...wish is 1.5J Right?? that would gave me 7.11 x10^-19 ..So why do i multiply 4.74 x 10^-19 by 1.5 jules if i alreday multiply that energy by the 3.16 x 10^-19 J to get the # of photons..???
     
    Last edited: Jan 3, 2008
  10. Jan 5, 2008 #9
    What i'm saying is that the energy you found is the amount of energy in a single photon, and it's a tiny number. Not even close to a whole joule. After the laser emits light for three seconds, you've measured a total energy of 1.5 joules. So if each photon makes a little tiny contribution, you need a whole lot of photons to make a whole joule... So what you need to multiply the energy by is the number of photons it takes to get a whole joule of energy. How do you do that?
     
  11. Jan 6, 2008 #10

    G01

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    Will.c is pointing out that your number of photons is a small fraction of one photon, which can't happen. 4.74 x 10^-19 is much less than one.

    I think you just made a typo with the sign in the exponent. I missed it as well. If you make your exponent positive that is the correct answer: 4.74 x 10^+19 photons should be what you get.
     
    Last edited: Jan 6, 2008
  12. Jan 6, 2008 #11
    ok..so i did..1photon has an energy of 3.16 x 10^-19J...how many photons in 1J ?
    (1J x 1photon)/3.16 x 10^-19J
    = 3.16 x 10^18 photons in 1 joul

    then if there are 3.16 x 10^18 photons in 1J..how many in 1.5 jouls(the total energy of the laser during the 3 second it was on)?
    (1.5J x 3.16 x 10^18 photons )/1J
    = 4.74 x 10^18 photons are emitted in a 3s interval...
    that would be the answer right...?
     
    Last edited: Jan 6, 2008
  13. Jan 6, 2008 #12

    G01

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    That work is correct.
     
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