91Ó°ÊÓ

The surface temperature is,$T=98.6°\text{ F}$

The surface area is,$A=4{\text{ c³¾}}^2$.

The wavelength range is, $\mathrm{Δ}\mathrm{λ}=1\text{ n³¾}$.

The expression of Wein’s displacement law to calculate the wavelength is given as follows.

${\mathrm{λ}}_{max}T=2898\text{‰ӾÀ.K}$ …(¾±)

The expression to calculate the spectral radiancy is given as follows.

$\mathit{S}\mathbf{\left(}\mathbf{λ}\mathbf{\right)}\mathbf{=}\frac{\mathbf{2}\mathbf{Ï€}{\mathbf{c}}^{\mathbf{2}}\mathbf{h}}{{\mathbf{λ}}^{\mathbf{5}}}\frac{\mathbf{1}}{{\mathbf{e}}^{\mathbf{h}\mathbf{c}\mathbf{/}\mathbf{λ}\mathbf{k}\mathbf{T}}\mathbf{−}\mathbf{1}}$ …(¾±¾±)

Here, his the plank’s constant, cis the speed of light, k is the Boltzmann constant and$\mathrm{λ}$is the wavelength.

The expression to calculate the radiated power is given as follows.

$\mathit{P}\mathbf{=}\mathit{S}\mathbf{\left(}\mathbf{λ}\mathbf{\right)}\mathit{A}\mathit{Δ}\mathit{λ}$ …(¾±¾±¾±)

Here,$\mathit{S}\mathbf{\left(}\mathbf{λ}\mathbf{\right)}$is the spectral radiancy.

The expression to calculate the rate of emitted photons per second is given as follows.

$\frac{\mathbf{d}\mathbf{N}}{\mathbf{d}\mathbf{t}}\mathbf{=}\frac{\mathbf{λ}\mathbf{P}}{\mathbf{h}\mathbf{c}}$ …(¾±±¹)

(a)

Calculate the temperature from Fahrenheit to kelvin,

$\begin{array}{l}T=(98.6\text{ F}−32)\text{ C}×\frac{5}{9}+273.15\text{‿é}\\ T=37\text{ C}+273.15\text{‿é}\\ T=310.15\text{‿é}\end{array}$

Calculate the wavelength,

Substitute$310.15\text{‿é}$ for T into equation (i).

$\begin{array}{l}{\mathrm{λ}}_{\max }×310.15\text{‿é}=2898\text{‰ӾÀ.K}\\ {\mathrm{λ}}_{\max }=\frac{2898\text{‰ӾÀ.K}}{310.15\text{‿é}}\\ {\mathrm{λ}}_{\max }=9.343\text{‰ӾÀ}\end{array}$

Hence the maximum value of wavelength at which spectral radiancy is maximum is $9.343\text{‰ӾÀ}$.

(b)

Calculate the spectral radiancy.

Substitute $1.38×{10}^{−23}\text{ J}/\text{K}$for K ,$6.62×{10}^{-34}\text{ }m\text{ }kg/s$ for h,$3×{10}^8\text{″¾}/{\text{c}}^{\text{2}}$ for c, $310.15\text{‿é}$for T and$9.343\text{‰ӾÀ}$ for$\mathrm{λ}$ into equation (i).

$\begin{array}{l}s\left(\mathrm{λ}\right)=\frac{2\mathrm{Ï€}{(3×{10}^8\text{″¾}/{\text{c}}^{\text{2}})}^2×6.62×{10}^{−34}\text{″¾kg}/\text{s}}{{(9.343×{10}^{−6}\text{″¾})}^5}×\frac{1}{e^{\frac{6.62×{10}^{−34}\text{″¾kg}/\text{s}×3×{10}^8\text{″¾}/{\text{c}}^{\text{2}}}{9.343×{10}^{−6}\text{″¾}×1.38×{10}^{−23}\text{ J}/\text{K}×310.15\text{‿é}}}−1}\\ s\left(\mathrm{λ}\right)=\frac{374.3521×{10}^{−18}{\text{″¾}}^{\text{3}}\text{.kg}/{\text{s.c}}^{\text{4}}}{7.119×{10}^{−26}{\text{″¾}}^{\text{5}}}×7.00×{10}^{−3}\\ s\left(\mathrm{λ}\right)=3.67×{10}^7\text{ W}/{\text{m}}^3\end{array}$

Calculate the radiated power.

Substitute$3.67×{10}^7\text{ W}/{\text{m}}^3$ for$S\left(\mathrm{λ}\right)$,$1\text{ n³¾}$ for $\mathrm{Δ}\mathrm{λ}$and$4{\text{ c³¾}}^2$ for A into equation (iii).

$\begin{array}{l}P=3.67×{10}^7\text{ W}/{\text{m}}^{\text{3}}×4×{10}^{−4}{\text{″¾}}^{\text{2}}×1×{10}^{−9}\text{″¾}\\ P=14.68×{10}^{−6}\text{ W}\\ P=1.468×{10}^{−5}\text{ W}\end{array}$

Hence the radiated power is $1.468×{10}^{−5}\text{ W}$.

(c)

Calculate the rate of emitted photons.

Substitute $1.468×{10}^{−5}\text{ W}$for P ,$9.343\text{‰ӾÀ}$ for $\mathrm{λ}$, $6.62×{10}^{-34}\text{ }m\text{ }kg/s$for h, and$3×{10}^8\text{″¾}/\text{s}$ for into equation (iv).

$\begin{array}{l}\frac{dN}{dt}=\frac{9.343×{10}^{−6}\text{″¾}×1.468×{10}^{−5}\text{ W}}{6.62×{10}^{−34}\text{″¾kg}/\text{s}×3×{10}^8\text{″¾}/\text{s}}\\ \frac{dN}{dt}=6.9×{10}^{14}\text{ p³ó´Ç³Ù´Ç²Ô²õ}/\text{s}\end{array}$

Hence the rate of emitted photons per second is $6.9×{10}^{14}\text{ p³ó´Ç³Ù´Ç²Ô²õ}/\text{s}$.

(d)

The value of the wavelength,$\mathrm{λ}=500\text{ n³¾}$

Calculate the spectral radiancy.

Substitute $1.38×{10}^{−23}\text{ J}/\text{K}$for k , $6.62×{10}^{-34}\text{ }m\text{ }kg/s$for h , $3×{10}^8\text{″¾}/\text{s}$for c,$310.15\text{‿é}$ for T and $500\text{ n³¾}$for$\mathrm{λ}$ into equation (i).

$\begin{array}{l}s\left(\mathrm{λ}\right)=\frac{2\mathrm{Ï€}{(3×{10}^8\text{″¾}/{\text{c}}^{\text{2}})}^2×6.62×{10}^{−34}\text{″¾kg}/\text{s}}{{(500×{10}^{−9}\text{″¾})}^5}×\frac{1}{e^{\frac{6.62×{10}^{−34}\text{″¾kg}/\text{s}×3×{10}^8\text{″¾}/{\text{c}}^{\text{2}}}{500×{10}^{−9}\text{″¾}×1.38×{10}^{−23}\text{ J}/\text{K}×310.15\text{‿é}}}−1}\\ s\left(\mathrm{λ}\right)=\frac{374.3521×{10}^{−18}{\text{″¾}}^{\text{3}}\text{.kg}/{\text{s.c}}^{\text{4}}}{3.125×{10}^{−32}{\text{″¾}}^{\text{5}}}×4.65×{10}^{−41}\\ s\left(\mathrm{λ}\right)=5.56×{10}^{−25}\text{ W}/{\text{m}}^3\end{array}$

Calculate the radiated power.

Substitute$5.56×{10}^{−25}\text{ W}/{\text{m}}^3$ for $S\left(\mathrm{λ}\right)$, $1\text{ n³¾}$for $\mathrm{Δ}\mathrm{λ}$and$4{\text{ c³¾}}^2$ for A into equation (iii).

$\begin{array}{l}P=5.56×{10}^{−25}\text{ W}/{\text{m}}^{\text{3}}×4×{10}^{−4}{\text{″¾}}^{\text{2}}×1×{10}^{−9}\text{″¾}\\ P=2.22×{10}^{−37}\text{ W}\end{array}$

Hence the radiated power is$2.22×{10}^{−37}\text{ W}$

(e)

Calculate the rate of emitted photons.

Substitute$2.22×{10}^{−37}\text{ W}$ for P, $500\text{ n³¾}$for $\mathrm{λ}$, $6.62×{10}^{-34}\text{ }m\text{ }kg/s$for h,and $3×{10}^8\text{″¾}/\text{s}$for c into equation (iv).

$\begin{array}{l}\frac{dN}{dt}=\frac{500×{10}^{−9}\text{″¾}×2.22×{10}^{−37}\text{ W}}{6.62×{10}^{−34}\text{″¾kg}/\text{s}×3×{10}^8\text{″¾}/\text{s}}\\ \frac{dN}{dt}=5.58×{10}^{−19}\text{ p³ó´Ç³Ù´Ç²Ô²õ}/\text{s}\end{array}$

Hence the rate of emitted photons per second is$5.58×{10}^{−19}\text{ p³ó´Ç³Ù´Ç²Ô²õ}/\text{s}$.