Question

The exitance (power per unit area per unit wavelength) from a blackbody (Box 20-1) is given by the Planck distribution: ${\mathbf{M}}_{\mathbf{\lambda }}\mathbf{=}\frac{\mathbf{2}{\mathbf{\pi hc}}^{\mathbf{2}}}{{\mathbf{\lambda }}^{\mathbf{5}}}\left(\frac{1}{e^{\mathrm{hc}/\mathrm{Akt}}-1}\right)$where $\mathit{\lambda }$is wavelength, Tis temperature K), his Planck’s constant, Cis the speed of light, and kis Boltzmann’s constant. The area under each curve between two wavelengths in the blackbody graph in Box 20-1is equal to the power per unit area$\left(W/m^2\right)$emitted between those two wavelengths. We find the area by integrating the Planck function between wavelengths${\mathit{\lambda }}_{\mathbf{1}}$and${\mathit{\lambda }}_{\mathbf{2}}$

role="math" localid="1664862982839" $\mathbf{Powe}\mathbf{}\mathbf{remitted}\mathbf{=}{\mathbf{\int }}_{{\mathbf{\lambda }}_{\mathbf{1}}}^{{\mathbf{\lambda }}_{\mathbf{2}}}{\mathbf{M}}_{\mathbf{\lambda }}\mathbf{d\lambda }$

For a narrow wavelength range, $\mathbf{∆}{\mathbf{\lambda }}_{\mathbf{1}}$the value of ${\mathbf{M}}_{\mathbf{\lambda }}$is nearly constant and the power emitted is simply the product ${\mathbf{M}}_{\mathbf{\lambda }}\mathbf{∆}{\mathbf{\lambda }}_{\mathbf{2}}$.

(a) Evaluate ${\mathbf{M}}_{\mathbf{h}}\mathbf{}\mathbf{at}\mathbf{}\mathbf{\lambda }\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\mu m}$and at$\mathbf{\lambda }\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{00}\mathbf{\mu m}\mathbf{}\mathbf{at}\mathbf{}\mathbf{T}\mathbf{=}\mathbf{1000}\mathbf{K}$

(b) Calculate the power emitted per square meter atin the intervalby evaluating the product

(c) Repeat part (b) for the interval$\mathbf{9}\mathbf{.}\mathbf{99}\mathbf{}\mathbf{to}\mathbf{}\mathbf{10}\mathbf{.}\mathbf{01}\mathbf{\mu m}$

(d) The quantity${\mathbf{M}}_{\mathbf{2}}{\mathbf{\mu }}_{\mathbf{m}}\mathbf{/}{\mathbf{M}}_{\mathbf{1}}\mathbf{0}\mathbf{\mu m}$is the relative exitance at the two wavelengths. Compare the relative exitance at these two wave-lengths atwith the relative exitance at 100K. What does your answer mean?

Short answer

a) ${\mathrm{M}}_{\mathrm{h}}\mathrm{at}\mathrm{\lambda }=2.00\mathrm{\mu m}$and at $\mathrm{\lambda }=10.00\mathrm{\mu m}\mathrm{at}\mathrm{T}=1000\mathrm{K}$

${\mathrm{M}}_{\mathrm{\lambda }}=8.79\cdot {10}^9\mathrm{W}/{\mathrm{m}}^3;{\mathrm{M}}_{\mathrm{\lambda }}=1.164\cdot {10}^9\mathrm{W}/{\mathrm{m}}^3$

b) The power emitted$175.8\mathrm{W}/{\mathrm{m}}^2$

c) The power emitted for the interval$9.99\mathrm{to}10.01\mathrm{\mu m}$

d) The relative exitance at the two wavelengths$7.55\cdot 3.17\cdot {10}^{-22}$

Step-by-step solution

Define planck’s law

Planck's law describes the spectral density of electromagnetic radiation emitted by a black body in thermal equilibrium at a given temperature T, when there is no net flow of matter or energy between the body and its environment.

Calculate the power emitted

a) The exitance is ${\mathrm{M}}_{\mathrm{\lambda }}=\frac{2{\mathrm{\pi hc}}^2}{{\mathrm{\lambda }}^5}.\frac{1}{{\mathrm{e}}^{{\displaystyle \frac{\mathrm{hc}}{\mathrm{kcT}}}-1}}$

$\mathrm{At}\mathrm{\lambda }=2\mathrm{\mu m},{\mathrm{theM}}_{\mathrm{\lambda }}\mathrm{is}:$

role="math" localid="1664864021991" $$\begin{gathered} {\mathrm{M}}_{\mathrm{\lambda }}=\frac{2\mathrm{\pi }6.626\cdot {10}^{-34}\mathrm{Js}\cdot (3\cdot {10}^8\mathrm{m}/{\mathrm{s}}^2{)}^2}{(2\cdot {10}^{-6}{\mathrm{m}}^2{)}^5}.\frac{1}{{\displaystyle \frac{6.626\cdot {10}^{-34}\mathrm{Js}\cdot 3\cdot {10}^8\mathrm{m}/{\mathrm{s}}^2}{{\mathrm{e}}^{2.{10}^{-6}}\cdot 1.38\cdot {10}^{-23}\mathrm{J}/\mathrm{K}\cdot 1000\mathrm{K})-1}}} \\ {\mathrm{M}}_{\mathrm{\lambda }}=8.79.{10}^9\mathrm{W}/{\mathrm{m}}^3 \end{gathered}$$

$\mathrm{At}\mathrm{\lambda }=10\mathrm{\mu m},\mathrm{the}{\mathrm{M}}_{\mathrm{\lambda }}\mathrm{is}:$

$$\begin{gathered} {\mathrm{M}}_{\mathrm{\lambda }}=\frac{2\mathrm{\pi }6.626\cdot {10}^{-34}\mathrm{Js}\cdot (3\cdot {10}^8\mathrm{m}/{\mathrm{s}}^2{)}^2}{(10.{10}^{-6}{\mathrm{m}}^2{)}^5}.\frac{1}{{\displaystyle \frac{6.626\cdot {10}^{-34}\mathrm{Js}\cdot 3\cdot {10}^8\mathrm{m}/{\mathrm{s}}^2}{{\mathrm{e}}^{10.{10}^{-6}}\cdot 1.38\cdot {10}^{-23}\mathrm{J}/\mathrm{K}\cdot 1000\mathrm{K})-1}}} \\ {\mathrm{M}}_{\mathrm{\lambda }}=1.164.{10}^9\mathrm{W}/{\mathrm{m}}^3 \end{gathered}$$

 Finding the power emitted

(b) The product

${\mathrm{M}}_{\mathrm{\lambda }}\mathrm{\Delta \lambda }\mathrm{is}$${\mathrm{M}}_{\mathrm{\lambda }}\mathrm{\Delta \lambda }=8.79\cdot {10}^9\mathrm{W}/{\mathrm{m}}^3\cdot 0.02\cdot {10}^{-6}\mathrm{m}=175.8\mathrm{W}/{\mathrm{m}}^2$

Finding the power emitted for the interval 9.99 to 10.01μm

(c) The product

${\mathrm{M}}_{\mathrm{\lambda }}∆\mathrm{\lambda }\mathrm{is}:$${\mathrm{M}}_{\mathrm{\lambda }}\mathrm{\Delta \lambda }=1.164\cdot {10}^9\mathrm{W}/{\mathrm{m}}^3\cdot 0.02\cdot {10}^{-6}\mathrm{m}=23.28\mathrm{W}/{\mathrm{m}}^2$

Calculating the relative existence

(d) $\mathrm{At}\mathrm{\lambda }=2\mathrm{\mu m}\mathrm{andT}=100\mathrm{K},\mathrm{the}{\mathrm{M}}_{\mathrm{\lambda }}\mathrm{is}:$

$$\begin{gathered} {\mathrm{M}}_{\mathrm{\lambda }}=\frac{2\mathrm{\pi }6.626\cdot {10}^{-34}\mathrm{Js}\cdot (3\cdot {10}^8\mathrm{m}/{\mathrm{s}}^2{)}^2}{(2\cdot {10}^{-6}{\mathrm{m}}^2{)}^5}.\frac{1}{{\displaystyle \frac{6.626\cdot {10}^{-34}\mathrm{Js}\cdot 3\cdot {10}^8\mathrm{m}/{\mathrm{s}}^2}{{\mathrm{e}}^{2.6.1.38.{10}^{-23}}\cdot \mathrm{J}/\mathrm{K}\cdot 1000\mathrm{K})}-1}} \\ {\mathrm{M}}_{\mathrm{\lambda }}=2..11.{10}^9\mathrm{W}/{\mathrm{m}}^3 \end{gathered}$$

Calculating the wavelength ofM2μm/M10μm  

$\mathrm{At}1000\mathrm{K}\mathrm{the}{\mathrm{M}}_{2\mathrm{\mu m}}/{\mathrm{M}}_{10\mathrm{\mu m}}\mathrm{is}:$

$$\begin{gathered} \frac{{\mathrm{M}}_{2\mathrm{\mu m}}}{{\mathrm{M}}_{10\mathrm{\mu m}}}=\frac{8.79.{10}^9\mathrm{W}/{\mathrm{m}}^3}{1.164.{10}^9\mathrm{W}/{\mathrm{m}}^3} \\ \frac{{\mathrm{M}}_{2\mathrm{\mu m}}}{{\mathrm{M}}_{10\mathrm{\mu m}}}7.55 \\ \mathrm{At}100\mathrm{K}\mathrm{the}{\mathrm{M}}_{2\mathrm{\mu m}}/{\mathrm{M}}_{10\mathrm{\mu m}}\mathrm{is}: \\ \frac{{\mathrm{M}}_{2\mathrm{\mu m}}}{{\mathrm{M}}_{10\mathrm{\mu m}}}=\frac{6.69.{10}^{-19}\mathrm{W}/{\mathrm{m}}^3}{2.11.{10}^3\mathrm{W}/{\mathrm{m}}^3} \\ \frac{{\mathrm{M}}_{2\mathrm{\mu m}}}{{\mathrm{M}}_{10\mathrm{\mu m}}}=3.17.{10}^{-22} \end{gathered}$$

We see that 1000K there is emission at both wavelengths, while 100K there is no emission at$2.00\mathrm{\mu m}$compared to$10.00\mathrm{\mu m}$.