Chapter 3, Problem 24P

(a)

To determine

The power of the emission in the visible region of the spectrum and compare it with ultraviolet and infrared region.

Answer to Problem 24P

The fraction of the visible power to the ultraviolet and infrared power is $0.073$ and $0.754$  respectively.

Explanation of Solution

Write the expression for the ratio of intensities from Planck’s law for the different wavelength.

  $\frac{l_1\left(\lambda ,t\right)}{l_2\left(\lambda ,t\right)}=\frac{\frac{2\pi h{c}^2}{{\lambda }_1^5}\frac{1}{\left[e^{hc/{\lambda }_{1}kT}-1\right]}}{\frac{2\pi h{c}^2}{{\lambda }_2^5}\frac{1}{\left[e^{hc/{\lambda }_{2}kT}-1\right]}}$        (I)

Here, $c$ is the speed of light, $h$ is the Planck’s constant, $k$ is the Boltzmann’s constant and $\lambda$ is the wavelength of emission.

Conclusion:

Substitute $400\text{nm}$ for ${\lambda }_1$, $966\text{nm}$ for ${\lambda }_2$, $1.38\times {10}^{-23}\text{J}/\text{K}$ for $k$, $1.986\times {10}^{-25}\text{J}\cdot \text{m}$ for $hc$ and $3000\text{K}$ for $T$ in equation (I).

  $\begin{array}{c}\frac{l_1\left(\lambda ,t\right)}{l_2\left(\lambda ,t\right)}=\frac{\frac{2\pi {\left(3\times {10}^8\text{m}/\text{s}\right)}^2\left(6.6\times {10}^{-34}\text{J}\cdot \text{s}\right)}{{\left(4\times {10}^{-7}\right)}^5}\frac{1}{\left[e^{\left(3\times {10}^8\text{m}/\text{s}\right)\left(6.6\times {10}^{-34}\text{J}\cdot \text{s}\right)/\left(4\times {10}^{-7}\right)\left(1.38\times {10}^{-23}\right)3000}-1\right]}}{\frac{2\pi {\left(3\times {10}^8\text{m}/\text{s}\right)}^2\left(6.6\times {10}^{-34}\text{J}\cdot \text{s}\right)}{{\left(9.66\times {10}^{-7}\right)}^5}\frac{1}{\left[e^{\left(3\times {10}^8\text{m}/\text{s}\right)\left(6.6\times {10}^{-34}\text{J}\cdot \text{s}\right)/\left(9.66\times {10}^{-7}\right)\left(1.38\times {10}^{-23}\right)3000}-1\right]}}\\ =0.073\end{array}$

Substitute $700\text{nm}$ for ${\lambda }_1$, $966\text{nm}$ for ${\lambda }_2$, $1.38\times {10}^{-23}\text{J}/\text{K}$ for $k$, $1.986\times {10}^{-25}\text{J}\cdot \text{m}$ for $hc$ and $3000\text{K}$ for $T$ in equation (I).

  $\begin{array}{c}\frac{l_1\left(\lambda ,t\right)}{l_2\left(\lambda ,t\right)}=\frac{\frac{2\pi {\left(3\times {10}^8\text{m}/\text{s}\right)}^2\left(6.6\times {10}^{-34}\text{J}\cdot \text{s}\right)}{{\left(4\times {10}^{-7}\right)}^5}\frac{1}{\left[e^{\left(3\times {10}^8\text{m}/\text{s}\right)\left(6.6\times {10}^{-34}\text{J}\cdot \text{s}\right)/\left(7\times {10}^{-7}\right)\left(1.38\times {10}^{-23}\right)3000}-1\right]}}{\frac{2\pi {\left(3\times {10}^8\text{m}/\text{s}\right)}^2\left(6.6\times {10}^{-34}\text{J}\cdot \text{s}\right)}{{\left(9.66\times {10}^{-7}\right)}^5}\frac{1}{\left[e^{\left(3\times {10}^8\text{m}/\text{s}\right)\left(6.6\times {10}^{-34}\text{J}\cdot \text{s}\right)/\left(9.66\times {10}^{-7}\right)\left(1.38\times {10}^{-23}\right)3000}-1\right]}}\\ =0.754\end{array}$

Thus, the fraction of the visible power to the ultraviolet and infrared power is $0.073$ and $0.754$  respectively.

(b)

To determine

The ratio of intensity at $400\text{nm}$ to $700\text{nm}$ wavelength with maximum intensity.

Explanation of Solution

Write the expression for the Planck’s law of radiation.

  $R=2\pi c^{2}h{\displaystyle {\int }_{{\lambda }_1}^{{\lambda }_2}{e}^{\left(-hc/\lambda kT\right)}{\lambda }^{-5}d\lambda }$        (II)

Here, $c$ is the speed of light, $h$ is the Planck’s constant, $k$ is the Boltzmann’s constant and $\lambda$ is the wavelength of emission.

Write the expression for the total power radiated by the lamp.

  $P=\sigma T^4$        (III)

Here, $P$ is the power emitted and $\sigma$ is the Stefan’s constant.

Write the expression for the fraction of power in the visible region.

  $r=\frac{R}{P}$        (IV)

Here, $r$ is the fraction of the power.

Conclusion:

Substitute $400\text{nm}$ for ${\lambda }_1$, $700\text{nm}$ for ${\lambda }_2$, $1.38\times {10}^{-23}\text{J}/\text{K}$ for $k$, $1.986\times {10}^{-25}\text{J}\cdot \text{m}$ for $hc$ and $3000\text{K}$ for $T$ in equation (II).

  $\begin{array}{c}R=2\pi {\left(3\times {10}^8\text{m}/\text{s}\right)}^2\left(6.6\times {10}^{-34}\text{J}\cdot \text{s}\right){\displaystyle {\int }_{4\times {10}^{-7}}^{7\times {10}^{-7}}{e}^{\left(-1.986\times {10}^{-25}\text{J}\cdot \text{m}/\lambda \left(4.14\times {10}^{-20}\right)\right)}{\lambda }^{-5}d\lambda }\\ =3.73\times {10}^{-16}{\displaystyle {\int }_{4\times {10}^{-7}}^{7\times {10}^{-7}}{e}^{\left(-1.986\times {10}^{-25}\text{J}\cdot \text{m}/\lambda \left(4.14\times {10}^{-20}\right)\right)}{\lambda }^{-5}d\lambda }\\ =3.71\times {10}^5\text{W}/{\text{m}}^{\text{2}}\end{array}$

Substitute $5.66\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}{\text{K}}^{\text{4}}$ for $\sigma$ and $3000\text{K}$ for $T$ in equation (III).

  $\begin{array}{c}P=5.66\times {10}^{-8}\text{W}/{\text{m}}^{\text{2}}{\text{K}}^{\text{4}}{\left(3000\text{K}\right)}^4\\ =4.59\times {10}^6\text{W}/{\text{m}}^{\text{2}}\end{array}$

Substitute $4.59\times {10}^6\text{W}/{\text{m}}^{\text{2}}$ for $P$ and $3.71\times {10}^5\text{W}/{\text{m}}^{\text{2}}$ for $R$ in equation (IV).

  $\begin{array}{c}r=\frac{3.71\times {10}^5\text{W}/{\text{m}}^{\text{2}}}{4.59\times {10}^6\text{W}/{\text{m}}^{\text{2}}}\\ =0.081\end{array}$

Thus, the fraction of the visible power to the maximum intensity power is $0.081$.