Which of the following statements is false regarding spectral distribution of thermal radiation emitted by a grey surface P. Monochromatic emissivity is independent of wavelength Q. Total emissivity is equal to monochromatic emissivity.R. (monochromatic emissive power) Vs λ( graph for the surface at agiven temperature is scaled down version of the graph for black body at same temperature.S. The wavelength corresponding to maximum monochromatic emissive power is same as that of grey surface at same temperature.a)Qb)Rc)Sd)None of theseCorrect answer is option 'D'. Can you explain this answer?

Question

Which of the following statements is false regarding spectral distribution of thermal radiation emitted by a grey surface P. Monochromatic emissivity is independent of wavelength Q. Total emissivity is equal to monochromatic emissivity.R. (monochromatic emissive power) Vs &lambda;( graph for the surface at agiven temperature is scaled down version of the graph for black body at same temperature.S. The wavelength corresponding to maximum monochromatic emissive power is same as that of grey surface at same temperature.a)Qb)Rc)Sd)None of theseCorrect answer is option 'D'. Can you explain this answer?

Answer

A grey surface is one for which monochromatic emissivity ε_λ is independent of wavelength (λ ) But total emissivity (ε) of a surface is given as

ε = E/E_b

Where E & E_brepresent total emissive power of the surface and a black surface at same temperature.

But as ε_λ ≠ f(λ)

∴ ε = ε_λE_b/E_b

∴ ε = ε_λ

i.e. total emissivity is equal to monochromatic emissivity. Hence statement Q is also correct.

In the figure shown,

AB = E_λ &

AC = E_bλ

∴ AB\Ac = E_λ/E_bλ

i.e, AB\AC = ε_λ

i.e. ratio of line segment AB and AC is monochromatic emissivity for surface corresponding to wavelength (λ) But as (ελ≠f(λ))

this ratio is same for all λ

i.e, A¹B¹/A¹C¹ = AB/AC = A²B²/A²C²

Thus graph of grey surface is simply scaled down version of graph of black surface. This also suggests that corresponding to maxima point for both surfaces is same.

Hence statement R & S are also correct.

Thus all statements are correct & option (D) is the correct answer.

Answer

A grey surface is one for which monochromatic emissivity ε_λ is independent of wavelength (λ ) But total emissivity (ε) of a surface is given as

ε = E/E_b

Where E & E_brepresent total emissive power of the surface and a black surface at same temperature.

But as ε_λ ≠ f(λ)

∴ ε = ε_λE_b/E_b

∴ ε = ε_λ

i.e. total emissivity is equal to monochromatic emissivity. Hence statement Q is also correct.

In the figure shown,

AB = E_λ &

AC = E_bλ

∴ AB\Ac = E_λ/E_bλ

i.e, AB\AC = ε_λ

i.e. ratio of line segment AB and AC is monochromatic emissivity for surface corresponding to wavelength (λ) But as (ελ≠f(λ))

this ratio is same for all λ

i.e, A¹B¹/A¹C¹ = AB/AC = A²B²/A²C²

Thus graph of grey surface is simply scaled down version of graph of black surface. This also suggests that corresponding to maxima point for both surfaces is same.

Hence statement R & S are also correct.

Thus all statements are correct & option (D) is the correct answer.

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