Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: I'm doing something wrong

  1. May 8, 2008 #1
    [SOLVED] I'm doing something wrong...

    1. A red laser with a wavelength of 650 nm and a blue laser with a wavelength of 450 nn emit laser beams with the same light power. How do their rates of photon emission compare? Answer this by computing R[tex]_{red}[/tex] / R[tex]_{blue}[/tex]



    2. P = Rhf = dNhf / dt = dE[tex]_{light}[/tex] / dt
    R = dN / dt
    E[tex]_{light}[/tex] = Nhf, where N is the number of photons
    f = c / [tex]\lambda[/tex]
    h = 6.624E-32 Js




    3. I have the f[tex]_{red}[/tex] = 4.62 x 10[tex]^{14}[/tex] s
    and f[tex]_{blue}[/tex] = 6.67 x 10[tex]^{14}[/tex] s , it says their Powers are the same, so I go ahead and went and equaled:

    P[tex]_{red}[/tex] to P[tex]_{blue}[/tex], which is:


    P[tex]_{red}[/tex]h[tex]_{red}[/tex]f[tex]_{red}[/tex] = P[tex]_{blue}[/tex]h[tex]_{blue}[/tex]f[tex]_{blue}[/tex]

    but I am tired and can't see the relevance, because when I multiplied times h [in Js] and then I divide like explained, I get .6925, and the ANSWER is 1.44.

    Help? Please?
     
  2. jcsd
  3. May 8, 2008 #2

    berkeman

    User Avatar

    Staff: Mentor

    Since red photons carry less energy per photon, there will need to be more of them. You just got your answer upside-down.

    Just use your equation for E = Nhf, and be sure to take the ratio in the correct direction.
     
  4. May 8, 2008 #3

    Chi Meson

    User Avatar
    Science Advisor
    Homework Helper

    Yes the answer is 1.44.

    You got confused because you plugged in your number way too soon. Consider this:

    Only the wavelengths are important. Obviously "R" here is used as "N," th number of photons per second? Well, P=P, you got that, and P=(Rhf)/t and f =c/ lamda.

    Just substitute, see what cancel out, and use what you are left with.
     
  5. May 8, 2008 #4
    I double checked the math, looks good.
     
  6. May 8, 2008 #5
    Ok....that was...really weird, I guess I somehow reversed wavelengths? I simply did the reciprocal and it worked, thanks all!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook