Question

Assuming that your surface temperature is$\mathbf{98}\mathbf{.6}\mathbf{°}\mathbf{\text{\hspace{0.17em}F}}$and that you are an ideal blackbody radiator (you are close), find

(a) the wavelength at which your spectral radiancy is maximum,

(b) the power at which you emit thermal radiation in a wavelength range of $\mathbf{1}\mathbf{.}\mathbf{00}\mathit{n}\mathit{m}$at that wavelength, from a surface area of$\mathbf{4}\mathbf{.00}{\mathbf{\text{\hspace{0.17em}cm}}}^{\mathbf{\text{2}}}$, and

(c) the corresponding rate at which you emit photons from that area. Using a wavelength of$\mathbf{500}\mathit{n}\mathit{m}$ (in the visible range),

(d) recalculate the power and

(e) the rate of photon emission. (As you have noticed, you do not visibly glow in the dark.)

Short answer

(a) The maximum value of wavelength at which spectral radiancy is maximum is,$9.343\text{\hspace{0.17em}μm}$

(b) The radiated power is, $1.468\times {10}^{-5}\text{\hspace{0.17em}W}$.

(c) The rate of emitted photons per second is,$6.9\times {10}^{14}\text{\hspace{0.17em}photons}/\text{s}$ .

(d)The radiated power is, $2.22\times {10}^{-37}\text{\hspace{0.17em}W}$.

(e) The rate of emitted photons per second is, $5.58\times {10}^{-19}\text{\hspace{0.17em}photons}/\text{s}$.

Step-by-step solution

Write the given data from the question.

The surface temperature is,$T=98.6°\text{\hspace{0.17em}F}$

The surface area is,$A=4{\text{\hspace{0.17em}cm}}^2$.

The wavelength range is, $\Delta \lambda =1\text{\hspace{0.17em}nm}$.

 Step 2: Determine the formulas to calculate wavelength, power and rate of emitting the photons from the surface. 

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

${\lambda }_{max}T=2898\text{\hspace{0.17em}μm.K}$ …(i)

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

$\mathit{S}\mathbf{\left(}\mathbf{\lambda }\mathbf{\right)}\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }{\mathbf{c}}^{\mathbf{2}}\mathbf{h}}{{\mathbf{\lambda }}^{\mathbf{5}}}\frac{\mathbf{1}}{{\mathbf{e}}^{\mathbf{h}\mathbf{c}\mathbf{/}\mathbf{\lambda }\mathbf{k}\mathbf{T}}\mathbf{-}\mathbf{1}}$ …(ii)

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

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

$\mathit{P}\mathbf{=}\mathit{S}\mathbf{\left(}\mathbf{\lambda }\mathbf{\right)}\mathit{A}\mathit{\Delta }\mathit{\lambda }$ …(iii)

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

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

$\frac{\mathbf{d}\mathbf{N}}{\mathbf{d}\mathbf{t}}\mathbf{=}\frac{\mathbf{\lambda }\mathbf{P}}{\mathbf{h}\mathbf{c}}$ …(iv)

Calculate the wavelength at which spectral radiancy is maximum

(a)

Calculate the temperature from Fahrenheit to kelvin,

$\begin{array}{l}T=(98.6\text{\hspace{0.17em}F}-32)\text{\hspace{0.17em}C}\times \frac{5}{9}+273.15\text{\hspace{0.17em}K}\\ T=37\text{\hspace{0.17em}C}+273.15\text{\hspace{0.17em}K}\\ T=310.15\text{\hspace{0.17em}K}\end{array}$

Calculate the wavelength,

Substitute$310.15\text{\hspace{0.17em}K}$ for T into equation (i).

$\begin{array}{l}{\lambda }_{\max }\times 310.15\text{\hspace{0.17em}K}=2898\text{\hspace{0.17em}μm.K}\\ {\lambda }_{\max }=\frac{2898\text{\hspace{0.17em}μm.K}}{310.15\text{\hspace{0.17em}K}}\\ {\lambda }_{\max }=9.343\text{\hspace{0.17em}μm}\end{array}$

Hence the maximum value of wavelength at which spectral radiancy is maximum is $9.343\text{\hspace{0.17em}μm}$.

 Step 4: Calculate the radiated power for the 1 nmwavelength range and 4 cm2surface area.

(b)

Calculate the spectral radiancy.

Substitute $1.38\times {10}^{-23}\text{\hspace{0.17em}J}/\text{K}$for K ,$6.62\times {10}^{-34}mkg/s$ for h,$3\times {10}^8\text{\hspace{0.17em}m}/{\text{c}}^{\text{2}}$ for c, $310.15\text{\hspace{0.17em}K}$for T and$9.343\text{\hspace{0.17em}μm}$ for$\lambda$ into equation (i).

$\begin{array}{l}s\left(\lambda \right)=\frac{2\pi {(3\times {10}^8\text{\hspace{0.17em}m}/{\text{c}}^{\text{2}})}^2\times 6.62\times {10}^{-34}\text{\hspace{0.17em}mkg}/\text{s}}{{(9.343\times {10}^{-6}\text{\hspace{0.17em}m})}^5}\times \frac{1}{e^{\frac{6.62\times {10}^{-34}\text{\hspace{0.17em}mkg}/\text{s}\times 3\times {10}^8\text{\hspace{0.17em}m}/{\text{c}}^{\text{2}}}{9.343\times {10}^{-6}\text{\hspace{0.17em}m}\times 1.38\times {10}^{-23}\text{\hspace{0.17em}J}/\text{K}\times 310.15\text{\hspace{0.17em}K}}}-1}\\ s\left(\lambda \right)=\frac{374.3521\times {10}^{-18}{\text{\hspace{0.17em}m}}^{\text{3}}\text{.kg}/{\text{s.c}}^{\text{4}}}{7.119\times {10}^{-26}{\text{\hspace{0.17em}m}}^{\text{5}}}\times 7.00\times {10}^{-3}\\ s\left(\lambda \right)=3.67\times {10}^7\text{\hspace{0.17em}W}/{\text{m}}^3\end{array}$

Calculate the radiated power.

Substitute$3.67\times {10}^7\text{\hspace{0.17em}W}/{\text{m}}^3$ for$S\left(\lambda \right)$,$1\text{\hspace{0.17em}nm}$ for $\Delta \lambda$and$4{\text{\hspace{0.17em}cm}}^2$ for A into equation (iii).

$\begin{array}{l}P=3.67\times {10}^7\text{\hspace{0.17em}W}/{\text{m}}^{\text{3}}\times 4\times {10}^{-4}{\text{\hspace{0.17em}m}}^{\text{2}}\times 1\times {10}^{-9}\text{\hspace{0.17em}m}\\ P=14.68\times {10}^{-6}\text{\hspace{0.17em}W}\\ P=1.468\times {10}^{-5}\text{\hspace{0.17em}W}\end{array}$

Hence the radiated power is $1.468\times {10}^{-5}\text{\hspace{0.17em}W}$.

Calculate the rate of emitted photons from the area.

(c)

Calculate the rate of emitted photons.

Substitute $1.468\times {10}^{-5}\text{\hspace{0.17em}W}$for P ,$9.343\text{\hspace{0.17em}μm}$ for $\lambda$, $6.62\times {10}^{-34}mkg/s$for h, and$3\times {10}^8\text{\hspace{0.17em}m}/\text{s}$ for into equation (iv).

$\begin{array}{l}\frac{dN}{dt}=\frac{9.343\times {10}^{-6}\text{\hspace{0.17em}m}\times 1.468\times {10}^{-5}\text{\hspace{0.17em}W}}{6.62\times {10}^{-34}\text{\hspace{0.17em}mkg}/\text{s}\times 3\times {10}^8\text{\hspace{0.17em}m}/\text{s}}\\ \frac{dN}{dt}=6.9\times {10}^{14}\text{\hspace{0.17em}photons}/\text{s}\end{array}$

Hence the rate of emitted photons per second is $6.9\times {10}^{14}\text{\hspace{0.17em}photons}/\text{s}$.

Calculate the radiated power for the wavelength range and surface area.

(d)

The value of the wavelength,$\lambda =500\text{\hspace{0.17em}nm}$

Calculate the spectral radiancy.

Substitute $1.38\times {10}^{-23}\text{\hspace{0.17em}J}/\text{K}$for k , $6.62\times {10}^{-34}mkg/s$for h , $3\times {10}^8\text{\hspace{0.17em}m}/\text{s}$for c,$310.15\text{\hspace{0.17em}K}$ for T and $500\text{\hspace{0.17em}nm}$for$\lambda$ into equation (i).

$\begin{array}{l}s\left(\lambda \right)=\frac{2\pi {(3\times {10}^8\text{\hspace{0.17em}m}/{\text{c}}^{\text{2}})}^2\times 6.62\times {10}^{-34}\text{\hspace{0.17em}mkg}/\text{s}}{{(500\times {10}^{-9}\text{\hspace{0.17em}m})}^5}\times \frac{1}{e^{\frac{6.62\times {10}^{-34}\text{\hspace{0.17em}mkg}/\text{s}\times 3\times {10}^8\text{\hspace{0.17em}m}/{\text{c}}^{\text{2}}}{500\times {10}^{-9}\text{\hspace{0.17em}m}\times 1.38\times {10}^{-23}\text{\hspace{0.17em}J}/\text{K}\times 310.15\text{\hspace{0.17em}K}}}-1}\\ s\left(\lambda \right)=\frac{374.3521\times {10}^{-18}{\text{\hspace{0.17em}m}}^{\text{3}}\text{.kg}/{\text{s.c}}^{\text{4}}}{3.125\times {10}^{-32}{\text{\hspace{0.17em}m}}^{\text{5}}}\times 4.65\times {10}^{-41}\\ s\left(\lambda \right)=5.56\times {10}^{-25}\text{\hspace{0.17em}W}/{\text{m}}^3\end{array}$

Calculate the radiated power.

Substitute$5.56\times {10}^{-25}\text{\hspace{0.17em}W}/{\text{m}}^3$ for $S\left(\lambda \right)$, $1\text{\hspace{0.17em}nm}$for $\Delta \lambda$and$4{\text{\hspace{0.17em}cm}}^2$ for A into equation (iii).

$\begin{array}{l}P=5.56\times {10}^{-25}\text{\hspace{0.17em}W}/{\text{m}}^{\text{3}}\times 4\times {10}^{-4}{\text{\hspace{0.17em}m}}^{\text{2}}\times 1\times {10}^{-9}\text{\hspace{0.17em}m}\\ P=2.22\times {10}^{-37}\text{\hspace{0.17em}W}\end{array}$

Hence the radiated power is$2.22\times {10}^{-37}\text{\hspace{0.17em}W}$

Calculate the rate of emitted photons from the area.

(e)

Calculate the rate of emitted photons.

Substitute$2.22\times {10}^{-37}\text{\hspace{0.17em}W}$ for P, $500\text{\hspace{0.17em}nm}$for $\lambda$, $6.62\times {10}^{-34}mkg/s$for h,and $3\times {10}^8\text{\hspace{0.17em}m}/\text{s}$for c into equation (iv).

$\begin{array}{l}\frac{dN}{dt}=\frac{500\times {10}^{-9}\text{\hspace{0.17em}m}\times 2.22\times {10}^{-37}\text{\hspace{0.17em}W}}{6.62\times {10}^{-34}\text{\hspace{0.17em}mkg}/\text{s}\times 3\times {10}^8\text{\hspace{0.17em}m}/\text{s}}\\ \frac{dN}{dt}=5.58\times {10}^{-19}\text{\hspace{0.17em}photons}/\text{s}\end{array}$

Hence the rate of emitted photons per second is$5.58\times {10}^{-19}\text{\hspace{0.17em}photons}/\text{s}$.