Chapter 7, Problem 7A.1P

(a)

Interpretation Introduction

Interpretation:

The energy density in the range $650\text{nm}$ to $655\text{nm}$ inside a cavity at $25°\text{C}$ has to be calculated.

Concept introduction:

Energy density is defined as the total energy inside the container divided by its volume.  The energy density at any temperature, $T$ due to the presence of the electromagnetic radiations of the wavelength ranging between $\lambda$ and $\lambda +\text{d}\lambda$ is known as energy spectral density.

Answer to Problem 7A.1P

The energy density in the range $650\text{nm}$ to $655\text{nm}$ inside a cavity at $25°\text{C}$ is $\underset{\_}{1.42\times 10^{-33}J{m}^{-3}}$.

Explanation of Solution

The formula for Planck’s distribution is,

    $\rho \left(\lambda ,T\right)=\frac{8\text{π}hc}{{\lambda }^5\left(e^{hc/\lambda kT}-1\right)}$                                                                               (1)

Where,

• $\lambda$ is the wavelength.
• $h$ is the Planck’s constant $\left(6.626\times {10}^{-34}\text{kg}{\text{m}}^2/\text{s}\right)$.
• $c$ is the speed of light $\left(3\times {10}^8\text{m}/\text{s}\right)$.
• $T$ is the temperature.
• $\rho$ is the energy spectral density.
• $k$ is the Boltzmann’s distribution constant $\left(1.38\times {10}^{-23}\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{-2}\right)$.

The range of wavelength ($\Delta \lambda$) is,

    $\begin{array}{c}\Delta \lambda =\text{655}\text{nm}-\text{650}\text{nm}\\ =\text{5}\text{nm}\end{array}$

The conversion of $\text{nm}$ to $\text{m}$ is done as,

  $\text{1}\text{nm}={10}^{-9}\text{m}$

Therefore, the conversion of $5\text{nm}$ to $\text{m}$ is done as,

  $5\text{nm}=5\times {10}^{-9}\text{m}$

The given range of wavelength is small.  Hence, the approximation can be used.

    $\Delta E=\rho \Delta \lambda$                                                                                                     (2)

Where,

• $\Delta E$ is the energy density.
• $\rho$ is the energy spectral density.
• $\Delta \lambda$ is the range of wavelength.

The value of $\lambda$ is calculated as,

    $\begin{array}{c}\lambda =\frac{655+650}{2}\\ =652.5\text{nm}\end{array}$

The conversion of $\text{nm}$ to $\text{m}$ is done as,

  $\text{1}\text{nm}={10}^{-9}\text{m}$

Therefore, the conversion of $652.5\text{nm}$ to $\text{m}$ is done as,

  $652.5\text{nm}=652.5\times {10}^{-9}\text{m}$

The value of $\lambda$ is $652.5\times {10}^{-9}\text{m}$.

Substitute the value of $h$, $c$, $k$ and $\lambda$ in equation (1).

    $\begin{array}{c}\rho \left(\lambda ,T\right)=\frac{8\times 3.14\times \left(6.626\times {10}^{-34}\text{kg}{\text{m}}^2/\text{s}\right)\times \left(3\times {10}^8\text{m}/\text{s}\right)}{{\left(652.5\times {10}^{-9}\text{m}\right)}^5\left(e^{6.626\times {10}^{-34}\times 3\times {10}^8/652.5\times {10}^{-9}\times 1.38\times {10}^{-23}\times T}-1\right)}\\ =\frac{4.99\times {10}^{-24}}{1.18\times {10}^{-31}\left(e^{2.2075\times {10}^4/T}-1\right)}\\ =\frac{4.2\times {10}^7}{\left(e^{2.2075\times {10}^4/T}-1\right)}\end{array}$

The given temperature is $25°\text{C}$.

The conversion of given temperature $\left(°\text{C}\right)$ into K is shown below.

  $\begin{array}{c}\text{K}=°\text{C+273}\\ =25°\text{C}+273\\ =298\text{K}\end{array}$

The value of $\Delta \lambda$ is $5\times {10}^{-9}\text{m}$.

The value of $\rho$ is $\frac{4.2\times {10}^7}{\left(e^{2.2075\times {10}^4/T}-1\right)}$.

The value of $T$ is $298\text{K}$.

Substitute the value of $\Delta \lambda$, $T$ and $\rho$ in equation (2).

    $\begin{array}{c}\Delta E=\frac{4.2\times {10}^7}{\left(e^{2.2075\times {10}^4/298}-1\right)}\times 5\times {10}^{-9}\text{m}\\ \text{=}\frac{4.2\times {10}^7}{\left(1.48\times {10}^{32}-1\right)}\times 5\times {10}^{-9}\\ =\frac{\text{0}\text{.21}}{1.48\times {10}^{32}}\\ =\underset{\_}{1.42\times 10^{-33}J{m}^{-3}}\end{array}$

Hence, the energy density in the range $650\text{nm}$ to $655\text{nm}$ inside a cavity at $25°\text{C}$ is $\underset{\_}{1.42\times 10^{-33}J{m}^{-3}}$.

(b)

Interpretation Introduction

Interpretation:

The energy density in the range $650\text{nm}$ to $655\text{nm}$ inside a cavity at $3000°\text{C}$ has to be calculated.

Concept introduction:

Energy density is defined as the total energy inside the container divided by its volume.  The energy density at any temperature, $T$ due to the presence of the electromagnetic radiations of the wavelength ranging between $\lambda$ and $\lambda +\text{d}\lambda$ is known as energy spectral density.

Answer to Problem 7A.1P

The energy density in the range $650\text{nm}$ to $655\text{nm}$ inside a cavity at $3000°\text{C}$ is $\underset{\_}{2.4\times 10^{-4}J{m}^{-3}}$.

Explanation of Solution

The formula for Planck’s distribution is,

    $\rho \left(\lambda ,T\right)=\frac{8\text{π}hc}{{\lambda }^5\left(e^{hc/\lambda kT}-1\right)}$                                                                               (1)

Where,

• $\lambda$ is the wavelength.
• $h$ is the Planck’s constant $\left(6.626\times {10}^{-34}\text{kg}{\text{m}}^2/\text{s}\right)$.
• $c$ is the speed of light $\left(3\times {10}^8\text{m}/\text{s}\right)$.
• $T$ is the temperature.
• $\rho$ is the energy spectral density.
• $k$ is the Boltzmann’s distribution constant $\left(1.38\times {10}^{-23}\text{kg}{\text{m}}^{\text{2}}{\text{s}}^{-2}\right)$.

The range of wavelength ($\Delta \lambda$) is,

    $\begin{array}{c}\Delta \lambda =\text{655}\text{nm}-\text{650}\text{nm}\\ =\text{5}\text{nm}\end{array}$

The conversion of $\text{nm}$ to $\text{m}$ is done as,

  $\text{1}\text{nm}={10}^{-9}\text{m}$

Therefore, the conversion of $5\text{nm}$ to $\text{m}$ is done as,

  $5\text{nm}=5\times {10}^{-9}\text{m}$

The given range of wavelength is small.  Hence, the approximation can be used.

    $\Delta E=\rho \Delta \lambda$                                                                                                     (2)

Where,

• $\Delta E$ is the energy density.
• $\rho$ is the energy spectral density.
• $\Delta \lambda$ is the range of wavelength.

The value of $\lambda$ is calculated as,

    $\begin{array}{c}\lambda =\frac{655+650}{2}\\ =652.5\text{nm}\end{array}$

The conversion of $\text{nm}$ to $\text{m}$ is done as,

  $\text{1}\text{nm}={10}^{-9}\text{m}$

Therefore, the conversion of $652.5\text{nm}$ to $\text{m}$ is done as,

  $652.5\text{nm}=652.5\times {10}^{-9}\text{m}$

The value of $\lambda$ is $652.5\times {10}^{-9}\text{m}$.

Substitute the value of $h$, $c$, $k$ and $\lambda$ in equation (1).

    $\begin{array}{c}\rho \left(\lambda ,T\right)=\frac{8\times 3.14\times \left(6.626\times {10}^{-34}\text{kg}{\text{m}}^2/\text{s}\right)\times \left(3\times {10}^8\text{m}/\text{s}\right)}{{\left(652.5\times {10}^{-9}\text{m}\right)}^5\left(e^{6.626\times {10}^{-34}\times 3\times {10}^8/652.5\times {10}^{-9}\times 1.38\times {10}^{-23}\times T}-1\right)}\\ =\frac{4.99\times {10}^{-24}}{1.18\times {10}^{-31}\left(e^{2.2075\times {10}^4/T}-1\right)}\\ =\frac{4.2\times {10}^7}{\left(e^{2.2075\times {10}^4/T}-1\right)}\end{array}$

The given temperature is $3000°\text{C}$.

The conversion of given temperature $\left(°\text{C}\right)$ into K is shown below.

  $\begin{array}{c}\text{K}=°\text{C+273}\\ =3000°\text{C}+273\\ =3273\text{K}\end{array}$

The value of $\Delta \lambda$ is $5\times {10}^{-9}\text{m}$.

The value of $\rho$ is $\frac{4.2\times {10}^7}{\left(e^{2.2075\times {10}^4/T}-1\right)}$.

The value of $T$ is $3273\text{K}$.

Substitute the value of $\Delta \lambda$, $T$ and $\rho$ in equation (2).

    $\begin{array}{c}\Delta E=\frac{4.2\times {10}^7}{\left(e^{2.2075\times {10}^4/3273}-1\right)}\times 5\times {10}^{-9}\text{m}\\ \text{=}\frac{4.2\times {10}^7}{\left(8.48\times {10}^2\right)}\times 5\times {10}^{-9}\\ =\frac{\text{0}\text{.21}}{\left(8.48\times {10}^2\right)}\\ =\underset{\_}{2.4\times 10^{-4}J{m}^{-3}}\end{array}$

Hence, the energy density in the range $650\text{nm}$ to $655\text{nm}$ inside a cavity at $3000°\text{C}$ is $\underset{\_}{2.4\times 10^{-4}J{m}^{-3}}$.