Number of photons radiated at cavity's hole

  • Thread starter jg370
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  • #1
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Homework Statement



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

Homework Equations



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

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

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

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

jg370
 

Answers and Replies

  • #2
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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.
 
  • #3
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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.
 
  • #4
ideasrule
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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.
 
  • #5
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jg370 said:
I have managed to perform the integration of Plank's wavelength distribution function by making the change of variable,
[tex] x= \frac{hc}{\lambda k_BT}[/tex]
and using the following integral:
[tex] \int \frac {x^3}{(e^x-1)} dx [/tex]

In order to use the above substitution, I have determined that [tex]\lambda[/tex] and [tex]d\lambda[/tex] as follows:

[tex]x=\frac{hc}{\lambdak_BT}, \lambda = \frac{hc}{xk_BT}, d\lambda = -\frac{hc}{x^2k_BT}[/tex]

I have work out the limit of integration to 3.83 and 3.84 using the following relation:
[tex] \frac{hc}{500*10^{-9}k_BT}[/tex] and [tex] \frac{hc}{501*10^{-9}k_BT}[/tex]

Consequently I came up with the following equation:

[tex]\frac{2\pi k_B^4T^4}{h^3c^2} \int_{3.83}^{3.84}\frac{x^3}{(e^x-1)}dx[/tex]

Evaluation of this equation yielded [tex]2.28 *10^{-28}[/tex] which is completely wrong, the correct answer being [tex]\frac{1.3*10^{15}}{s}[/tex]

Hopefully, someone will be able to point out where I went wrong.

Thanks for your help

jg370

 
  • #6
ideasrule
Homework Helper
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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
[tex]
\frac{2 k_B^6T^4}{h^3c^2}
[/tex]

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.
 
  • #7
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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?
 
  • #8
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I get the correct order of magnitude, but not the right answer:
[tex]
\frac{2hc^{2}}{\lambda^{5}}\frac{1}{e^{\frac{hc}{{\lambda}kT}}-1}\Delta\lambda{\Omega}A
[/tex]
So (using 500nm for lambda)
[tex]
\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
[/tex]

Multiplied together gives .00208W and

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

Maybe someone can see where I'm going wrong?
 
  • #9
ideasrule
Homework Helper
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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.
 
  • #10
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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)
 
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  • #11
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jg370 said:
Following some research, I found reference to the radiation power from a cavity hole. This reference is from Michael Fowler of University of West Virginia and can be found at:

http://galileo.phys.virginia.edu/classes/252/black_body_radiation.html

Here goes:

1.The energy density inside an oven, denoted by [tex]\rho(f,T)[/tex] means that at temperature T, the energy in Joules/m^3 in the frequency interval [tex] f, f+\Delta f[/tex] is:

[tex]\rho(f,T) \Delta f[/tex]

2. To find the power radiated out of the hole, bear in mind that the radiation inside the oven has waves equally going both ways—so only half of them will come out through the hole. Also, if the hole has area A, waves coming from the inside at an angle will see a smaller target area.

The result of these two effects is that the radiation power from hole area A is:

[tex] \frac{1}{4} Ac\rho(f,T)[/tex]

Hopefully, this will help to get closure on my problem.

Also, I have the following question: why is it that integrating Plank's law by means of the change of variable
[tex]x=\frac{hc}{\lambda kT}[/tex]

and the performing the following operation

[tex]\frac{2\pi k^4T^4}{h^3c^2}\int_{\lambda_2}^{\lambda1} \frac{x^3}{e^x -1}[/tex]

while taking care to determine the limits of integration [tex]\lambda1,\lambda 2[/tex],according to the afore mentionned change of variable does not work out.

Please note that I did proceed as such and so indicated in one of my earlier reply.

Thanks for your help

 

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