Chapter 40, Problem 12P

(a)

To determine

The power radiated by the black body.

Answer to Problem 12P

The power radiated by the black body is $70,870\text{W}$.

Explanation of Solution

Write the expression for the power radiated by the black body.

    $P=\sigma Ae{T}^4$                                                                                                         (I)

Here, $P$ is the power radiated by the black body, $A$ is the area, $\sigma$ is the Stefan-Boltzmann constant, $e$ is the emissivity and $T$ is the temperature.

Conclusion:

Substitute $5.6696\times {10}^{-8}{\text{W/m}}^2\cdot {\text{K}}^4$ for $\sigma$, $20.0{\text{cm}}^2$ for $A$, $1$ for $e$ and $5000\text{K}$ for $T$ in equation (I) to find $P$.

    $\begin{array}{c}P=\left(5.6696\times {10}^{-8}{\text{W/m}}^2\cdot {\text{K}}^4\right)\left(20.0{\text{cm}}^2\right)\left(1\right){\left(5000\text{K}\right)}^4\\ =\left(5.6696\times {10}^{-8}{\text{W/m}}^2\cdot {\text{K}}^4\right)\left(\left(20.0{\text{cm}}^2\right)\left(\frac{{10}^{-4}{\text{m}}^2}{1{\text{cm}}^2}\right)\right)\left(1\right){\left(5000\text{K}\right)}^4\\ =70,870\text{W}\end{array}$

Thus, the power radiated by the black body is $70,870\text{W}$.

(b)

To determine

The wavelength at which the blackbody radiate most intensely.

Answer to Problem 12P

The wavelength at which the blackbody radiate most intensely is $580\text{nm}$.

Explanation of Solution

Write the equation for the wavelength at which the blackbody radiate most intensely.

    ${\lambda }_{\max }=\frac{2.898\times {10}^{-3}\text{m}\cdot \text{K}}{T}$                                                                        (II)

Here, ${\lambda }_{\max }$ is the maximum wavelength.

Conclusion:

Substitute $5000\text{K}$ for $T$ in equation (II) to find ${\lambda }_{\max }$.

    $\begin{array}{c}{\lambda }_{\max }=\frac{2.898\times {10}^{-3}\text{m}\cdot \text{K}}{5000\text{K}}\\ =\left(5.796\times {10}^{-7}\text{m}\right)\left(\frac{{10}^{-9}\text{nm}}{1\text{m}}\right)\\ \approx 580\text{nm}\end{array}$

Thus, the wavelength at which the blackbody radiate most intensely is $580\text{nm}$.

(c)

To determine

The spectral power per wavelength interval at $580\text{nm}$.

Answer to Problem 12P

The spectral power per wavelength interval at $580\text{nm}$ is $8.0\times {10}^{10}\text{W/m}$.

Explanation of Solution

Write the equation for $C$.

    $C=\frac{hc}{k_{B}T}$                                                                                             (III)

Here, $C$ is the constant, $h$ is the Plank’s constant, $c$ is the speed of light and $k_B$ is the Boltzmann constant.

Write the equation for $D$.

    $D=2\pi h{c}^{2}A$                                                                                        (IV)

Here, $D$ is a constant.

Write the equation for the power per wavelength interval.

    $P\left(\lambda \right)=\frac{2\pi h{c}^{2}A}{{\lambda }^5\left[\exp \left(hc/\lambda k_{B}T\right)-1\right]}$

Substitute $C$ for $\frac{hc}{k_{B}T}$ and $D$ for $2\pi h{c}^{2}A$.

    $P\left(\lambda \right)=\frac{D}{{\lambda }^5\left[\exp \left(C/\lambda \right)-1\right]}$                                                                            (V)

Conclusion:

Substitute $6.626\times {10}^{-34}{\text{m}}^2\cdot \text{kg/s}$ for $h$, $3\times {10}^8\text{m/s}$ for $c$, $1.38\times {10}^{-23}\text{J/K}$ for $k_B$ and $5000\text{K}$ for $T$ in equation (III) to find $C$.

    $\begin{array}{c}C=\frac{\left(6.626\times {10}^{-34}{\text{m}}^2\cdot \text{kg/s}\right)\left(3\times {10}^8\text{m/s}\right)}{\left(1.38\times {10}^{-23}\text{J/K}\right)\left(5000\text{K}\right)}\\ =0.00288\times {10}^{-3}\text{m}\\ =2.88\times {10}^{-6}\text{m}\end{array}$

Substitute $6.626\times {10}^{-34}{\text{m}}^2\cdot \text{kg/s}$ for $h$, $3\times {10}^8\text{m/s}$ for $c$ and $20.0{\text{cm}}^2$ for $T$ in equation (IV) to find $C$.

    $\begin{array}{c}D=2\pi \left(6.626\times {10}^{-34}{\text{m}}^2\cdot \text{kg/s}\right){\left(3\times {10}^8\text{m/s}\right)}^2\left(20.0{\text{cm}}^2\right)\\ =2\left(3.14\right)\left(6.626\times {10}^{-34}{\text{m}}^2\cdot \text{kg/s}\right){\left(3\times {10}^8\text{m/s}\right)}^2\left(\left(20.0{\text{cm}}^2\right)\left(\frac{{10}^{-4}{\text{m}}^2}{1{\text{cm}}^2}\right)\right)\\ =7.50\times {10}^{-19}\text{J}\cdot {\text{m}}^2\text{/s}\end{array}$

Substitute $2.88\times {10}^{-6}\text{m}$ for $C$, $7.50\times {10}^{-19}\text{J}\cdot {\text{m}}^2\text{/s}$ for $D$ and $580\text{nm}$ for $\lambda$ in equation (V) to find $P\left(\lambda \right)$.

    $\begin{array}{c}P\left(\lambda \right)=\frac{7.50\times {10}^{-19}\text{J}\cdot {\text{m}}^2\text{/s}}{{\left(580\text{nm}\right)}^5\left[\exp \left(\frac{2.88\times {10}^{-6}\text{m}}{580\text{nm}}\right)-1\right]}\\ =\frac{7.50\times {10}^{-19}\text{J}\cdot {\text{m}}^2\text{/s}}{{\left(\left(580\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\right)}^5\left[\exp \left(\frac{2.88\times {10}^{-6}\text{m}}{\left(580\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\right)-1\right]}\\ =7.99\times {10}^{10}\text{W/m}\\ \simeq \text{8}\text{.0}\times {10}^{10}\text{W/m}\end{array}$

Thus, the spectral power per wavelength interval at $580\text{nm}$ is $8.0\times {10}^{10}\text{W/m}$.

(d)

To determine

The spectral power per wavelength interval at $1.00\text{nm}$.

Answer to Problem 12P

The spectral power per wavelength interval at $1\text{nm}$ is $9.42\times {10}^{-1226}\text{W/m}$.

Explanation of Solution

From the equation (V) in part (c), the spectral power per wavelength is.

    $P\left(\lambda \right)=\frac{D}{{\lambda }^5\left[\exp \left(C/\lambda \right)-1\right]}$

Conclusion:

Substitute $2.88\times {10}^{-6}\text{m}$ for $C$, $7.50\times {10}^{-19}\text{J}\cdot {\text{m}}^2\text{/s}$ for $D$ and $1.00\text{nm}$ for $\lambda$ in above equation to find $P\left(\lambda \right)$.

$\begin{array}{c}P\left(\lambda \right)=\frac{7.50\times {10}^{-19}\text{J}\cdot {\text{m}}^2\text{/s}}{{\left(1\text{nm}\right)}^5\left[\exp \left(\frac{2.88\times {10}^{-6}\text{m}}{1\text{nm}}\right)-1\right]}\\ =\frac{7.50\times {10}^{-19}\text{J}\cdot {\text{m}}^2\text{/s}}{{\left(\left(1\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\right)}^5\left[\exp \left(\frac{2.88\times {10}^{-6}\text{m}}{\left(1\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\right)-1\right]}\\ =9.42\times {10}^{-1226}\text{W/m}\end{array}$

Thus, the spectral power per wavelength interval at $1\text{nm}$ is $9.42\times {10}^{-1226}\text{W/m}$.

(e)

To determine

The spectral power per wavelength interval at $5.00\text{nm}$.

Answer to Problem 12P

The spectral power per wavelength interval at $5.00\text{nm}$ is $1.00\times {10}^{-227}\text{W/m}$.

Explanation of Solution

From the equation (V) in part (c), the spectral power per wavelength is.

    $P\left(\lambda \right)=\frac{D}{{\lambda }^5\left[\exp \left(C/\lambda \right)-1\right]}$

Conclusion:

Substitute $2.88\times {10}^{-6}\text{m}$ for $C$, $7.50\times {10}^{-19}\text{J}\cdot {\text{m}}^2\text{/s}$ for $D$ and $5.00\text{nm}$ for $\lambda$ in above equation to find $P\left(\lambda \right)$.

    $\begin{array}{c}P\left(\lambda \right)=\frac{7.50\times {10}^{-19}\text{J}\cdot {\text{m}}^2\text{/s}}{{\left(5\text{nm}\right)}^5\left[\exp \left(\frac{2.88\times {10}^{-6}\text{m}}{5\text{nm}}\right)-1\right]}\\ =\frac{7.50\times {10}^{-19}\text{J}\cdot {\text{m}}^2\text{/s}}{{\left(\left(5\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\right)}^5\left[\exp \left(\frac{2.88\times {10}^{-6}\text{m}}{\left(5\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\right)-1\right]}\\ =1.00\times {10}^{-227}\text{W/m}\end{array}$

Thus, the spectral power per wavelength interval at $5.00\text{nm}$ is $1.00\times {10}^{-227}\text{W/m}$.

(f)

To determine

The spectral power per wavelength interval at $400\text{nm}$.

Answer to Problem 12P

The spectral power per wavelength interval at $400\text{nm}$ is $5.44\times {10}^{10}\text{W/m}$.

Explanation of Solution

From the equation (V) in part (c), the spectral power per wavelength is.

    $P\left(\lambda \right)=\frac{D}{{\lambda }^5\left[\exp \left(C/\lambda \right)-1\right]}$

Conclusion:

Substitute $2.88\times {10}^{-6}\text{m}$ for $C$, $7.50\times {10}^{-19}\text{J}\cdot {\text{m}}^2\text{/s}$ for $D$ and $400\text{nm}$ for $\lambda$ in above equation to find $P\left(\lambda \right)$.

    $\begin{array}{c}P\left(\lambda \right)=\frac{7.50\times {10}^{-19}\text{J}\cdot {\text{m}}^2\text{/s}}{{\left(400\text{nm}\right)}^5\left[\exp \left(\frac{2.88\times {10}^{-6}\text{m}}{400\text{nm}}\right)-1\right]}\\ =\frac{7.50\times {10}^{-19}\text{J}\cdot {\text{m}}^2\text{/s}}{{\left(\left(400\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\right)}^5\left[\exp \left(\frac{2.88\times {10}^{-6}\text{m}}{\left(400\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\right)-1\right]}\\ =5.44\times {10}^{10}\text{W/m}\end{array}$

Thus, the spectral power per wavelength interval at $400\text{nm}$ is $5.44\times {10}^{10}\text{W/m}$.

(g)

To determine

The spectral power per wavelength interval at $700\text{nm}$.

Answer to Problem 12P

The spectral power per wavelength interval at $700\text{nm}$ is $7.38\times {10}^{10}\text{W/m}$.

Explanation of Solution

From the equation (V) in part (c), the spectral power per wavelength is.

    $P\left(\lambda \right)=\frac{D}{{\lambda }^5\left[\exp \left(C/\lambda \right)-1\right]}$

Conclusion:

Substitute $2.88\times {10}^{-6}\text{m}$ for $C$, $7.50\times {10}^{-19}\text{J}\cdot {\text{m}}^2\text{/s}$ for $D$ and $700\text{nm}$ for $\lambda$ in above equation to find $P\left(\lambda \right)$.

    $\begin{array}{c}P\left(\lambda \right)=\frac{7.50\times {10}^{-19}\text{J}\cdot {\text{m}}^2\text{/s}}{{\left(700\text{nm}\right)}^5\left[\exp \left(\frac{2.88\times {10}^{-6}\text{m}}{700\text{nm}}\right)-1\right]}\\ =\frac{7.50\times {10}^{-19}\text{J}\cdot {\text{m}}^2\text{/s}}{{\left(\left(700\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\right)}^5\left[\exp \left(\frac{2.88\times {10}^{-6}\text{m}}{\left(700\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\right)-1\right]}\\ =7.38\times {10}^{10}\text{W/m}\end{array}$

Thus, the spectral power per wavelength interval at $700\text{nm}$ is $7.38\times {10}^{10}\text{W/m}$.

(h)

To determine

The spectral power per wavelength interval at $1.00\text{mm}$.

Answer to Problem 12P

The spectral power per wavelength interval at $1.00\text{mm}$ is $0.260\text{W/m}$.

Explanation of Solution

From the equation (V) in part (c), the spectral power per wavelength is.

    $P\left(\lambda \right)=\frac{D}{{\lambda }^5\left[\exp \left(C/\lambda \right)-1\right]}$

Conclusion:

Substitute $2.88\times {10}^{-6}\text{m}$ for $C$, $7.50\times {10}^{-19}\text{J}\cdot {\text{m}}^2\text{/s}$ for $D$ and $1.00\text{mm}$ for $\lambda$ in above equation to find $P\left(\lambda \right)$.

    $\begin{array}{c}P\left(\lambda \right)=\frac{7.50\times {10}^{-19}\text{J}\cdot {\text{m}}^2\text{/s}}{{\left(1.00\text{mm}\right)}^5\left[\exp \left(\frac{2.88\times {10}^{-6}\text{m}}{1.00\text{mm}}\right)-1\right]}\\ =\frac{7.50\times {10}^{-19}\text{J}\cdot {\text{m}}^2\text{/s}}{{\left(\left(1.00\text{mm}\right)\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right)}^5\left[\exp \left(\frac{2.88\times {10}^{-6}\text{m}}{\left(1.00\text{mm}\right)\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}\right)-1\right]}\\ =0.260\text{W/m}\end{array}$

Thus, the spectral power per wavelength interval at $1.00\text{mm}$ is $0.260\text{W/m}$.

(I)

To determine

The spectral power per wavelength interval at $10.0\text{cm}$.

Answer to Problem 12P

The spectral power per wavelength interval at $10.0\text{cm}$ is $2.60\times {10}^{-9}\text{W/m}$.

Explanation of Solution

From the equation (V) in part (c), the spectral power per wavelength is.

    $P\left(\lambda \right)=\frac{D}{{\lambda }^5\left[\exp \left(C/\lambda \right)-1\right]}$

Conclusion:

Substitute $2.88\times {10}^{-6}\text{m}$ for $C$, $7.50\times {10}^{-19}\text{J}\cdot {\text{m}}^2\text{/s}$ for $D$ and $10.0\text{cm}$ for $\lambda$ in above equation to find $P\left(\lambda \right)$.

    $\begin{array}{c}P\left(\lambda \right)=\frac{7.50\times {10}^{-19}\text{J}\cdot {\text{m}}^2\text{/s}}{{\left(10.0\text{cm}\right)}^5\left[\exp \left(\frac{2.88\times {10}^{-6}\text{m}}{10.0\text{cm}}\right)-1\right]}\\ =\frac{7.50\times {10}^{-19}\text{J}\cdot {\text{m}}^2\text{/s}}{{\left(\left(10.0\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^5\left[\exp \left(\frac{2.88\times {10}^{-6}\text{m}}{\left(10.0\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}\right)-1\right]}\\ =2.60\times {10}^{-9}\text{W/m}\end{array}$

Thus, the spectral power per wavelength interval at $10.0\text{cm}$ is $2.60\times {10}^{-9}\text{W/m}$.

(J)

To determine

The power radiated by the object as visible light.

Answer to Problem 12P

The power radiated by the object as visible light is $19\text{kW}$.

Explanation of Solution

The wavelengths $400\text{nm}$ and $700\text{nm}$ comes in the visible region. Thus, when taking power, the average power at these wavelengths are used.

Write the equation for the power radiated.

    $P=P\left(\overline{\lambda }\right)\Delta \lambda$ (VI)

Here, $P$ is the radiated power in the visible region, $P\left(\overline{\lambda }\right)$ is the average power in the visible area and $\Delta \lambda$ is the change in wavelength.

Conclusion:

The average power is calculated from the visible area.

  $\begin{array}{c}P\left(\overline{\lambda }\right)=\frac{5.44\times {10}^{10}\text{W/m}+7.38\times {10}^{10}\text{W/m}}{2}\\ =6.41\times {10}^{10}\text{W/m}\end{array}$

Substitute $6.41\times {10}^{10}\text{W/m}$ for $P\left(\overline{\lambda }\right)$ and $700\text{nm}-400\text{nm}$ for $\Delta \lambda$ in equation (VI) to find $P$.

    $\begin{array}{c}P=\left(6.41\times {10}^{10}\text{W/m}\right)\left(700\text{nm}-400\text{nm}\right)\\ =\left(6.41\times {10}^{10}\text{W/m}\right)\left(\left(300\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\right)\\ =\left(1.92\times {10}^4\text{W}\right)\left(\frac{{10}^{-3}\text{kW}}{1\text{W}}\right)\\ =19\text{kW}\end{array}$

Thus, the power radiated by the object as visible light is $19\text{kW}$.