91Ó°ÊÓ

The exitance (power per unit area per unit wavelength) from a blackbody (Box 20-1) is given by the Planck distribution: ${\mathbf{M}}_{\mathbf{λ}}\mathbf{=}\frac{\mathbf{2}{\mathbf{Ï€³ó³¦}}^{\mathbf{2}}}{{\mathbf{λ}}^{\mathbf{5}}}\left(\frac{1}{e^{\mathrm{hc}/\mathrm{Akt}}-1}\right)$where $\mathit{λ}$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${\mathit{λ}}_{\mathbf{1}}$and${\mathit{λ}}_{\mathbf{2}}$

role="math" localid="1664862982839" $\mathbf{Powe}\mathbf{}\mathbf{remitted}\mathbf{=}{\mathbf{∫}}_{{\mathbf{λ}}_{\mathbf{1}}}^{{\mathbf{λ}}_{\mathbf{2}}}{\mathbf{M}}_{\mathbf{λ}}\mathbf{»åλ}$

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

(a) Evaluate ${\mathbf{M}}_{\mathbf{h}}\mathbf{}\mathbf{at}\mathbf{}\mathbf{λ}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{μ³¾}$and at$\mathbf{λ}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{00}\mathbf{μ³¾}\mathbf{}\mathbf{at}\mathbf{}\mathbf{T}\mathbf{=}\mathbf{1000}\mathbf{K}$

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

(c) Repeat part (b) for the interval$\mathbf{9}\mathbf{.}\mathbf{99}\mathbf{}\mathbf{to}\mathbf{}\mathbf{10}\mathbf{.}\mathbf{01}\mathbf{μ³¾}$

(d) The quantity${\mathbf{M}}_{\mathbf{2}}{\mathbf{μ}}_{\mathbf{m}}\mathbf{/}{\mathbf{M}}_{\mathbf{1}}\mathbf{0}\mathbf{μ³¾}$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?

Step by step solution

01

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.

02

Calculate the power emitted

a) The exitance is ${\mathrm{M}}_{\mathrm{λ}}=\frac{2{\mathrm{Ï€³ó³¦}}^2}{{\mathrm{λ}}^5}.\frac{1}{{\mathrm{e}}^{{\displaystyle \frac{\mathrm{hc}}{\mathrm{kcT}}}-1}}$

$\mathrm{At}\mathrm{λ}=2\mathrm{μ³¾},{\mathrm{theM}}_{\mathrm{λ}}\mathrm{is}:$

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

$\mathrm{At}\mathrm{λ}=10\mathrm{μ³¾},\mathrm{the}{\mathrm{M}}_{\mathrm{λ}}\mathrm{is}:$

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

03

 Finding the power emitted

(b) The product

${\mathrm{M}}_{\mathrm{λ}}\mathrm{Δλ}\mathrm{is}$${\mathrm{M}}_{\mathrm{λ}}\mathrm{Δλ}=8.79â‹\dots {10}^9\mathrm{W}/{\mathrm{m}}^3â‹\dots 0.02â‹\dots {10}^{-6}\mathrm{m}=175.8\mathrm{W}/{\mathrm{m}}^2$

04

Finding the power emitted for the interval 9.99 to 10.01μm

(c) The product

${\mathrm{M}}_{\mathrm{λ}}∆\mathrm{λ}\mathrm{is}:$${\mathrm{M}}_{\mathrm{λ}}\mathrm{Δλ}=1.164â‹\dots {10}^9\mathrm{W}/{\mathrm{m}}^3â‹\dots 0.02â‹\dots {10}^{-6}\mathrm{m}=23.28\mathrm{W}/{\mathrm{m}}^2$

05

Calculating the relative existence

(d) $\mathrm{At}\mathrm{λ}=2\mathrm{μ³¾}\mathrm{andT}=100\mathrm{K},\mathrm{the}{\mathrm{M}}_{\mathrm{λ}}\mathrm{is}:$

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

06

Calculating the wavelength ofM2μm/M10μm  

$\mathrm{At}1000\mathrm{K}\mathrm{the}{\mathrm{M}}_{2\mathrm{μ³¾}}/{\mathrm{M}}_{10\mathrm{μ³¾}}\mathrm{is}:$

$$\begin{gathered} \frac{{\mathrm{M}}_{2\mathrm{μ³¾}}}{{\mathrm{M}}_{10\mathrm{μ³¾}}}=\frac{8.79.{10}^9\mathrm{W}/{\mathrm{m}}^3}{1.164.{10}^9\mathrm{W}/{\mathrm{m}}^3} \\ \frac{{\mathrm{M}}_{2\mathrm{μ³¾}}}{{\mathrm{M}}_{10\mathrm{μ³¾}}}7.55 \\ \mathrm{At}100\mathrm{K}\mathrm{the}{\mathrm{M}}_{2\mathrm{μ³¾}}/{\mathrm{M}}_{10\mathrm{μ³¾}}\mathrm{is}: \\ \frac{{\mathrm{M}}_{2\mathrm{μ³¾}}}{{\mathrm{M}}_{10\mathrm{μ³¾}}}=\frac{6.69.{10}^{-19}\mathrm{W}/{\mathrm{m}}^3}{2.11.{10}^3\mathrm{W}/{\mathrm{m}}^3} \\ \frac{{\mathrm{M}}_{2\mathrm{μ³¾}}}{{\mathrm{M}}_{10\mathrm{μ³¾}}}=3.17.{10}^{-22} \end{gathered}$$

We see that 1000K there is emission at both wavelengths, while 100K there is no emission at$2.00\mathrm{μ³¾}$compared to$10.00\mathrm{μ³¾}$.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!