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Number of photons radiated at cavity's hole

  1. Dec 15, 2009 #1
    1. The problem statement, all variables and given/known data

    A blackhole at 7500K consist of an opening of diameter 0.0500mm, looking into the oven. Find the number of photons per second escaping the hole and having wavelengths between 500nm and 501nm

    2. Relevant equations

    [tex] P = \sigma AeT^4[/tex]

    [tex] Watt= \frac{Joule}{sec}[/tex]

    [tex] E_{p} = \frac{hc}{\lambda}[/tex]

    3. The attempt at a solution

    The total power radiated at the hole is:

    [tex] P= 5.67 * 10^{-8} \frac{w}{m^2 k^4}* 1.96*10^{-9} m^2 * (7500K)^4 = 0.35 w[/tex]

    I have equated the above answer (0.35 w) according to the following relationship:

    [tex] watt = \frac {joule}{sec}[/tex]

    Therefore the total energy radiated at the hole is [tex] 0.35 \frac{j}{s}[/tex]

    Energy of a photon at the cavity's hole is given by:

    [tex] E_p = \frac {hc}{\lambda} = \frac {6.626 *10^{34} js * 3 *10^8 \frac{m}{s}}{500 * 10^{-9}} = 3.97 * 10^{-19} j [/tex]

    The number of photons with wavelength 500 nm escaping the hole is:

    [tex] N_p = \frac {0.35 \frac{j}{s}}{3.97*10^{-19} j} = \frac{880 *10^{15}}{s}[/tex]

    This is not the correct answer; my textbook gives [tex] 1.3*10^{15}[/tex]

    I think it my be wrong to equate [tex] w = \frac{joule}{s}[/tex] and I am uncertain what to do about the statement "having wavelengths between 500nm and 501nm" in the question.

    Hopefully someone will provide indication(s) that will allow me to get to the correct answer.

    Thank you kindly

  2. jcsd
  3. Dec 15, 2009 #2
    You probably need to take into account Planck's Law. That is, .35W is the total power radiated by the blackbody, but only a portion of that is in the 500-501nm range, so you need to integrate over that part of the spectrum.
  4. Dec 18, 2009 #3
    Then, why does the problem states a radius for the cavity"s hole. Plank's law does not specify a specific area for the radia tion intensity.
  5. Dec 18, 2009 #4


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    It doesn't specify an area for radiation intensity because intensity is measured in W/m^2. To get from that to watts, you need to multiply by the area.
  6. Dec 20, 2009 #5
  7. Dec 20, 2009 #6


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    First, [tex]
    \frac{2\pi k_B^4T^4}{h^3c^2} \int_{3.83}^{3.84}\frac{x^3}{(e^x-1)}dx
    [/tex] isn't 3.83; it's 3.83*10^-19. Ditto for the other boundary.

    Also, check the constant in your final integration equation. I think it should be
    \frac{2 k_B^6T^4}{h^3c^2}

    I doubt, though, that you need to use integrals for this question. Just evaluate [tex]I'(\lambda,T) =\frac{2 hc^2}{\lambda^5}\frac{1}{ e^{\frac{hc}{\lambda kT}}-1}[/tex], multiply the result by 1 nm (since the difference from 500 to 501 nm is 1 nm), and multiply by the radius of the hole. The "I" in Planck's law isn't going to vary much from 500 to 501 nm.
  8. Dec 21, 2009 #7
    Thank you for your suggestions.

    I have verfied that the constant that I used is probably correct. Moreover, the limit of integration that I derived are based on the wavelengths expresed in nm, as you can see in my last post.

    Nevertheless, I carried out the calculation again and came up with the following answer:

    [tex] 2.4*10^5 \frac{J}{m^2s^2}[/tex]

    which is quite different from my textbook answer as mentionned in my last post.

    I feel that the units of the above answer indicate that something is seriously amist.

    I have also tried to follow the second method that you did suggest, and I got an even more nonsensical answer.

    Is it time to give up?
  9. Dec 21, 2009 #8
    I get the correct order of magnitude, but not the right answer:
    So (using 500nm for lambda)
    \frac{2hc^{2}}{\lambda^{5}}=3.81E15 \frac{kg}{m*s^3} , \frac{1}{e^{\frac{hc}{{\lambda}kT}}-1}=.0222 ,
    \Delta\lambda=10^{-9}m ,
    \Omega=4\pi ,
    A=1.96*10^{-9} m^2

    Multiplied together gives .00208W and

    \frac{.00208 W}{\frac{hc}{\lambda}} = 5.24 * 10^{15} Hz

    Maybe someone can see where I'm going wrong?
  10. Dec 21, 2009 #9


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    Where did you get [itex]{\Omega}A[/itex], and how did you calculate A? [itex] \frac{2hc^{2}}{\lambda^{5}}\frac{1}{e^{\frac{hc}{{ \lambda}kT}}-1}\Delta\lambda[/itex] has units of W/m^2, so all you have to do is multiply by the area of the hole, which is pi*r^2.
  11. Dec 21, 2009 #10
    According to [URL [Broken][/URL] I' has units of J/s/m^2/sr/m so I used the A = area of the hole and Omega = 4*pi steradians (the solid angle of a sphere)
    Last edited by a moderator: May 4, 2017
  12. Dec 23, 2009 #11
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