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Thermal Properties of Matter — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsThermal Properties of Matter3 Marks
Question
Two bodies A and B have thermal emissivity of 0.01 and 0.81 respectively. The outer surface area as of both the bodies are same. The two emit the same total radiated power. The wavelength $\lambda_\text{B}$ corresponding to the maximum intensity in radiations from B, is shifted from the wavelength $\lambda_\text{A}$ a corresponding to the maximum intensity in radiations from A by $1\mu\text{m}.$ If the temperature of body A is 5802K, find the temperature of body B and the wavelength ago $\lambda_\text{B}.$
✓

Answer

Here, $\varepsilon_\text{A}=0.01$ and $\varepsilon_\text{B}=0.81$ $\Big(\frac{\text{dQ}}{\text{dt}}\Big)_\text{A}=\Big(\frac{\text{dQ}}{\text{dt}}\Big)_\text{B}$ $\therefore\varepsilon_\text{A}\sigma_\text{A}\text{T}^\text{4}_\text{A}=\varepsilon_\text{B}\sigma_\text{A}\text{T}^4_\text{B}\Rightarrow\frac{\varepsilon_\text{A}}{\varepsilon_\text{B}}=\frac{\text{T}^4_\text{B}}{\text{T}^4_\text{A}}$ But $\lambda\text{T}=$ Contant, $\therefore\lambda_\text{A}\text{T}_\text{A}=\lambda_\text{B}\text{T}_\text{B}$ $\therefore\frac{\varepsilon_\text{A}}{\varepsilon_\text{B}}=\frac{\lambda^4_\text{A}}{\lambda^4_\text{B}}=\frac{0.1}{0.81}=\Big(\frac{1}{3}\Big)^4$ $\therefore\frac{\lambda_\text{A}}{\lambda_\text{B}}=\frac{1}{3}$ $\Rightarrow\lambda_\text{B}=3\lambda_\text{A}...(1)$ But $\lambda_\text{B}-\lambda_\text{A}=10^{-6}...(2)$ $\therefore3\lambda_\text{A}-\lambda_\text{A}=10^{-6}$ $\therefore\lambda_\text{A}=0.5\times10^{-6}\text{m}$ and $\lambda_\text{B}=3\times0.5\times10^{-6}$ $=1.5\times10^{-6}\text{m}$

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