Solved Problem on

A cavity radiator at 6000 K has a 0.10 mm diameter hole drilled into its wall. Find the power radiated through the hole in the wavelength range from 5500 Å to 5510 Å.

Problem data:
  • Cavity temperature:    T = 6000 K;
  • Hole diameter:    d = 0.10 mm;
  • Minimum wavelength:    λ1 = 5500 Å;
  • Maximum wavelength:    λ2 = 5510 Å.
  • Speed ​​of light:    c = 2.998 ×108 m/s;
  • Planck Constant::    h = 6.63 ×10--34 J.s;
  • Boltzmann Constant::    k = 1.38 ×10--23 J/K.

a) The total radiance is defined as the radiated power per unit area
\[ \begin{gathered} R_{T}=\frac{P}{A} \end{gathered} \]
the integral of the radiance over all frequencies gives the radiated power
\[ \begin{gather} P=A\int {}R_{T}(\nu)\;d\nu \tag{I} \end{gather} \]
The relationship between spectral radiance and energy density is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {R_{T}(\nu)=\frac{c}{4}\rho (\nu)} \tag{II} \end{gather} \]
substituting expression (II) into expression (I) and integrating it into the frequency interval of the problem
\[ \begin{gather} P=A\int_{\nu_{1}}^{\nu_{2}}{}\frac{c}{4}\rho (\nu)\;d\nu \\[5pt] P=\frac{Ac}{4}\int_{\nu_{1}}^{\nu_{2}}{}\rho (\nu)\;d\nu \tag{III} \end{gather} \]
Planck's Radiation Law for energy density is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {\rho (\nu)\;d\nu =\frac{8\pi \nu^{3}}{c^{3}}\frac{h}{\operatorname{e}^{h\nu/{kT}}-1}\;d\nu} \tag{IV} \end{gather} \]
The relationship between energy density, as a function of frequency, and energy, as a function of wavelength, is given by
\[ \begin{gather} \rho_{T}(\lambda )=-\rho_{T}(\nu)\;\frac{d\nu }{d\lambda } \tag{V} \end{gather} \]
The frequency is given by
\[ \begin{gather} \bbox[#99CCFF,10px] {\nu =\frac{c}{\lambda}} \tag{VI} \end{gather} \]
the derivative of frequency with respect to wavelength is given by
\[ \begin{gather} \frac{d\nu}{d\lambda}=\frac{d(c\lambda^{-1})}{d\lambda}=-c\lambda^{-2}=-{\frac{c}{\lambda^{2}}} \tag{VII} \end{gather} \]
substituting the expressions (IV), (VI), and (VII) into expression (V)
\[ \begin{gathered} \rho_{T}(\lambda)=-\left[\frac{8\pi}{c^{3}}\left(\frac{c}{\lambda}\right)^{3}\frac{h}{\operatorname{e}^{hc/{kT\lambda}}-1}\right]\;\left(-{\frac{c}{\lambda^{2}}\;d\lambda}\right)\\[5pt] \rho_{T}(\lambda)=\frac{8\pi c^{3}}{c^{3}\lambda^{3}}\frac{c}{\lambda^{2}}\frac{h}{\operatorname{e}^{hc/{kT\lambda}}-1}\;d\lambda \end{gathered} \]
\[ \begin{gathered} \bbox[#99CCFF,10px] {\rho_{T}(\lambda)\;d\lambda =\frac{8\pi c}{\lambda^{5}}\frac{h}{\operatorname{e}^{hc/{kT\lambda }}-1}\;d\lambda} \end{gathered} \]
Expression (III) can be written as
\[ \begin{gathered} P=\frac{Ac}{4}\int_{\lambda_{1}}^{\lambda_{2}}{}\frac{8\pi c}{\lambda ^{5}}\frac{h}{\operatorname{e}^{hc/{kT\lambda}}-1}\;d\lambda \end{gathered} \]
Note: We don't need to calculate the integral, we can approximate the value using the Mean Value Theorem for Integrals.
The integral of a function f(x) in an interval [a, b] represents the area under the curve (Figure 1), by the Mean Value Theorem for Integrals we have a value f(c) of the function that determines a rectangle based on equal to the length of the gap and height f(c) (Figure 2).

Figure 1
Figure 2
Point c is located anywhere in the interval [a, b], such that the value f(c) gives us equal areas under the curves.
In particular, if the function f(x) is linear the point c is at the midpoint of the interval [a, b] (Figure 3). This happens because the areas above and below the value of f(c) compensate each other.

Figure 3

In the problem, the difference in wavelengths relative to the given values ​​is
\[ \begin{gathered} \frac{\Delta \lambda }{\lambda _{1}}=\frac{\lambda _{2}-\lambda_{1}}{\lambda_{1}}=\frac{5510\times 10^{-10}-5500\times 10^{-10}}{5500\times 10^{-10}}=0.002 \end{gathered} \]
\[ \begin{gathered} \frac{\Delta \lambda }{\lambda _{2}}=\frac{\lambda _{2}-\lambda_{1}}{\lambda_{2}}=\frac{5510\times 10^{-10}-5500\times 10^{-10}}{5510\times 10^{-10}}=0.002 \end{gathered} \]
we see that the variation in the range of wavelengths varies from 2 parts per 1000, this range is small (Figure 4), we can approximate the infinitesimal range of wavelengths, , by the range, Δλ, and the variable of integration, λ, can be substituted by its average value, λm (Figure 5).

Figure 4
Figure 5

We can substitute the integral to the power with the following expression
\[ \begin{gathered} P=\frac{Ac}{4}\frac{8\pi c}{\lambda_{m}^{5}}\frac{h}{\operatorname{e}^{hc/{kT\lambda_{m}}}-1}\;\Delta \lambda \\[5pt] P=\frac{2A\pi c^{2}}{\lambda_{m}^{5}}\frac{h}{\operatorname{e}^{hc/{kT\lambda_{m}}}-1}\;\Delta \lambda \end{gathered} \]
Substituting the integration variable λ by the average value of the wavelength range
\[ \begin{gathered} \lambda_{m}=\frac{\lambda_{1}+\lambda_{2}}{2}\\[5pt] \lambda_{m}=\frac{5500\times 10^{-10}+5510\times 10^{-10}}{2}\\[5pt] \lambda_{m}=\frac{11010\times 10^{-10}}{2}\\[5pt] \lambda_{m}=5505\times 10^{-10}\;\overset{\circ}{\text{A}} \end{gathered} \]
and substituting the differential interval dλ by the interval Δλ given by
\[ \begin{gathered} \Delta \lambda =\lambda_{1}-\lambda_{2}\\[5pt] \Delta\lambda =5510\times 10^{-10}-5500\times 10^{-10}\\[5pt] \Delta \lambda=10\times 10^{-10}\;\overset{\circ}{\text{A}} \end{gathered} \]
The area of the hole will be
\[ \begin{gathered} A=\pi r^{2}\\[5pt] A=\pi\left(\frac{d}{2}\right)^{2}\\[5pt] A=3.14\times \left(\frac{10\times 10^{-3}}{2}\right)^{2}\\[5pt] A=7.85\times 10^{-5}\;\text{m}^{2} \end{gathered} \]
substituting the data
\[ \begin{align} P=& \frac{2\times (7.85\times 10^{-5})\times 3.14\times (2.998\times 10^{8})^{2}}{(5505\times 10^{-10})^{5}}\times\\ &\times{\frac{(6.63\times 10^{-34})}{\operatorname{e}^{(6.63\times 10^{-34})\times (2.998\times 10^{8})/{[(1.38\times 10^{-23})\times (6000)\times 5505\times 10^{-10}]}}-1}\times (10\times 10^{-10})}\;\Delta\lambda\\[5pt] &\qquad \qquad \quad P=\frac{4.43\times 10^{13}}{5.06\times 10^{-32}}\times \frac{6.63\times 10^{-34}}{\operatorname{e}^{4.36}-1}\times (10\times 10^{-10})\\[5pt] &\qquad \qquad \qquad P=8.75\times 10^{44}\times 8.58\times 10^{-36}\times 10\times 10^{-10} \end{align} \]
\[ \begin{gathered} \bbox[#FFCCCC,10px] {P=7.5\;\text{W}} \end{gathered} \]