Question

rcise 2 ZCT 1042025 A.pdf 6 / 13 104\% 5. Assuming that your surface temperature is \( 98.6^{\circ} \mathrm{F} \) and that you are an ideal blackbody radiator (you are close), find (i) the wavelength at which your spectral radiancy is maximum. (ii) the power at which you emit thermal radiation in a wavelength range of 1.00 nm at that wavelength, from a surface area of \( 4.00 \mathrm{~cm}^{2} \). (iii) the corresponding rate at which you emit photons from that area. Using a wavelength of 500 nm (in the visible range), (iv) recalculate the power (v) the rate of photon emission. Ans: T in Kelvin: \[ \mathrm{T}=\left(98.6^{\circ} \mathrm{F}-32^{\circ} \mathrm{F}\right)\left(\frac{{ }^{\frac{5}{9}} K}{{ }^{9} \mathrm{~F}}\right)+273.15=310.15 K \] (i) \( \lambda_{\text {max }}=\frac{b}{T}=\frac{2.898 \times 10^{-3} m \cdot K}{310.15 K}=9.34 \mu m \) (ii) Given: \( \Delta \lambda=1.00 \mathrm{~nm}, \mathrm{~A}=4.00 \mathrm{~cm}^{2}=4 \times 10^{-4} \mathrm{~m}^{2} \) We know that, intensity (I) can be written as: \[ \begin{aligned} I & =\frac{P}{A} \\ I(T) & =\int_{0}^{\infty} R(\lambda, T) d \lambda \end{aligned} \] Search

          rcise 2 ZCT 1042025 A.pdf
6 / 13
104\%
5. Assuming that your surface temperature is \( 98.6^{\circ} \mathrm{F} \) and that you are an ideal blackbody radiator (you are close), find
(i) the wavelength at which your spectral radiancy is maximum.
(ii) the power at which you emit thermal radiation in a wavelength range of 1.00 nm at that wavelength, from a surface area of \( 4.00 \mathrm{~cm}^{2} \).
(iii) the corresponding rate at which you emit photons from that area.

Using a wavelength of 500 nm (in the visible range),
(iv) recalculate the power
(v) the rate of photon emission.

Ans:
T in Kelvin:
\[
\mathrm{T}=\left(98.6^{\circ} \mathrm{F}-32^{\circ} \mathrm{F}\right)\left(\frac{{ }^{\frac{5}{9}} K}{{ }^{9} \mathrm{~F}}\right)+273.15=310.15 K
\]
(i) \( \lambda_{\text {max }}=\frac{b}{T}=\frac{2.898 \times 10^{-3} m \cdot K}{310.15 K}=9.34 \mu m \)
(ii) Given: \( \Delta \lambda=1.00 \mathrm{~nm}, \mathrm{~A}=4.00 \mathrm{~cm}^{2}=4 \times 10^{-4} \mathrm{~m}^{2} \)

We know that, intensity (I) can be written as:
\[
\begin{aligned}
I & =\frac{P}{A} \\
I(T) & =\int_{0}^{\infty} R(\lambda, T) d \lambda
\end{aligned}
\]
Search

rcise 2 ZCT 1042025 A.pdf
6 / 13
104%
5. Assuming that your surface temperature is 98.6^∘F and that you are an ideal blackbody radiator (you are close), find
(i) the wavelength at which your spectral radiancy is maximum.
(ii) the power at which you emit thermal radiation in a wavelength range of 1.00 nm at that wavelength, from a surface area of 4.00 cm^2.
(iii) the corresponding rate at which you emit photons from that area.

Using a wavelength of 500 nm (in the visible range),
(iv) recalculate the power
(v) the rate of photon emission.

Ans:
T in Kelvin:

T=(98.6^∘F-32^∘F)((^(5)/(9) K)/(^9 F))+273.15=310.15 K

(i) λmax =(b)/(T)=(2.898 × 10^-3 m · K)/(310.15 K)=9.34 μ m
(ii) Given: Δλ=1.00 nm, A=4.00 cm^2=4 × 10^-4 m^2

We know that, intensity (I) can be written as:

I =(P)/(A)

I(T) =∫0^∞ R(λ, T) d λ

Search

Added by Darnell C.

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