Chapter 9, Problem 9.32E

Calculate the power of light in the wavelength range $\lambda =350\text{nm}-351\text{nm}$ (that is, let $d\lambda$ be $\Delta \lambda =1\text{nm}$ in the Planck’s law, and let $\lambda$ be $350.5\text{nm}$) at temperatures of $1000\text{K}$, $3000\text{K}$ and $10,000\text{K}$.

Interpretation Introduction

Interpretation:

The power of light in the wavelength range $\lambda =350-351\text{nm}$ at temperatures $1000\text{K}$, $3000\text{K}$ and $10,000\text{K}$ is to be calculated.

Concept introduction:

Planck’s equation can also be represented in the form of power flux or infinitesimal power per unit area for a black body radiation at a given temperature and wavelengths as shown below.

$d\xi =\frac{2\pi h{c}^2}{{\lambda }^5}\left(\frac{1}{e^{hc/\lambda kT}-1}\right)d\lambda$

Answer to Problem 9.32E

The power of light in the wavelength range $\lambda =350-351\text{nm}$ at the temperatures $1000\text{K}$, $3000\text{K}$ and $10,000\text{K}$ is $1.01\times {10}^{-10}{\text{W/m}}^{\text{2}}$, $80.21{\text{W/m}}^{\text{2}}$ and $1.18\times {10}^6{\text{W/m}}^{\text{2}}$ respectively.

Explanation of Solution

It is given that wavelength range is $350\text{nm}-351\text{nm}$, the value of $d\lambda$ is $1\text{nm}$ and $\lambda$ is $350.5\text{nm}$.

The formula to calculate the power of light is,

$d\xi =\frac{2\pi h{c}^2}{{\lambda }^5}\left(\frac{1}{e^{hc/\lambda kT}-1}\right)d\lambda$

Where,

• $\xi$ is the power flux.

• $\lambda$ is the wavelength.

• $h$ is the Planck’s constant.

• $k$ is the Boltzmann constant.

• $c$ is the speed of light.

Substitute the values of constants and temperature $1000\text{K}$ to calculate the power of light in the given formula.

$\begin{array}{rll}d\xi & =& \left(\begin{array}{c}\frac{2\left(3.14\right)\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right){\left(3\times {10}^8{\text{m s}}^{-1}\right)}^2}{{\left(350.5\times {\text{10}}^{-9}\text{ m}\right)}^5}\left(\frac{1}{e^{\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 3\times {10}^8{\text{ms}}^{-1}}{350.5\times {10}^{-9}\text{m}\times 1.38\times {10}^{-23}{\text{JK}}^{\text{-1}}\times 1000\text{K}}}-1}\right)\\ \times \left(1\times {10}^{-9}\text{m}\right)\end{array}\right)\\ & =& \frac{3.745\times {10}^{-16}}{5.289\times {10}^{-33}}\times \left(\frac{1}{e^{\frac{1.9878\times {10}^{-25}}{4.8369\times {10}^{-27}}}-1}\right)\times 1\times {10}^{-9}\text{ J}\cdot {\text{s}}^{-1}{\text{m}}^{-2}\\ & =& \frac{3.745\times {10}^{-16}}{5.289\times {10}^{-33}}\times \frac{1}{7.00\times {10}^{17}-1}\times 1\times {10}^{-9}\text{ J}\cdot {\text{s}}^{-1}{\text{m}}^{-2}\\ & =& 7.08\times {10}^{16}\times 1.428\times {10}^{-18}\times 1\times {10}^{-9}\text{ J}\cdot {\text{s}}^{-1}{\text{m}}^{-2}\end{array}$

On simplifying the above equation,

$\text{1 W}=\text{1 J}\cdot {\text{s}}^{-1}$

$d\xi =1.01\times {10}^{-10}{\text{W/m}}^{\text{2}}$

Thus, the value of power flux is $1.01\times {10}^{-10}{\text{W/m}}^{\text{2}}$.

Substitute the values of constants and temperature $3000\text{K}$ to calculate the power of light in the given formula.

$\begin{array}{rll}d\xi & =& \left(\begin{array}{l}\frac{2\left(3.14\right)\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right){\left(3\times {10}^8{\text{ms}}^{-1}\right)}^2}{{\left(350.5\times {\text{10}}^{-9}\right)}^5}\left(\frac{1}{e^{\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 3\times {10}^8{\text{ms}}^{-1}}{350.5\times {10}^{-9}\text{m}\times 1.38\times {10}^{-23}{\text{JK}}^{\text{-1}}\times 3000\text{K}}}-1}\right)\\ \left(1\times {10}^{-9}\text{m}\right)\end{array}\right)\\ & =& \frac{3.745\times {10}^{-16}}{5.289\times {10}^{-33}}\times \left(\frac{1}{e^{\frac{1.9878\times {10}^{-25}}{1.451\times {10}^{-26}}}-1}\right)\times 1\times {10}^{-9}\text{ J}\cdot {\text{s}}^{-1}{\text{m}}^{-2}\\ & =& \frac{3.745\times {10}^{-16}}{5.289\times {10}^{-33}}\times \frac{1}{8.8\times {10}^5-1}\times 1\times {10}^{-9}\text{ J}\cdot {\text{s}}^{-1}{\text{m}}^{-2}\\ & =& 7.08\times {10}^{16}\times 1.133\times {10}^{-6}\times 1\times {10}^{-9}\text{ J}\cdot {\text{s}}^{-1}{\text{m}}^{-2}\end{array}$

On simplifying the above equation,

$\text{1 W}=\text{1 J}\cdot {\text{s}}^{-1}$

$d\xi =1.01\times {10}^{-10}{\text{W/m}}^{\text{2}}$

Thus, the value of power flux is $80.21{\text{W/m}}^{\text{2}}$.

Substitute the values of constants and temperature $10,000\text{K}$ to calculate the power of light in the given formula.

$\begin{array}{rll}d\xi & =& \left(\begin{array}{l}\frac{2\left(3.14\right)\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right){\left(3\times {10}^8{\text{ms}}^{-1}\right)}^2}{{\left(350.5\times {\text{10}}^{-9}\right)}^5}\left(\frac{1}{e^{\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}\times 3\times {10}^8{\text{ms}}^{-1}}{350.5\times {10}^{-9}\text{m}\times 1.38\times {10}^{-23}{\text{JK}}^{\text{-1}}\times 10000\text{K}}}-1}\right)\\ \left(1\times {10}^{-9}\text{m}\right)\end{array}\right)\\ & =& \frac{3.745\times {10}^{-16}}{5.289\times {10}^{-33}}\times \left(\frac{1}{e^{\frac{1.9878\times {10}^{-25}}{4.8369\times {10}^{-26}}}-1}\right)\times 1\times {10}^{-9}\text{ J}\cdot {\text{s}}^{-1}{\text{m}}^{-2}\\ & =& \frac{3.745\times {10}^{-16}}{5.289\times {10}^{-33}}\times \frac{1}{60.88-1}\times 1\times {10}^{-9}\text{ J}\cdot {\text{s}}^{-1}{\text{m}}^{-2}\\ & =& 7.08\times {10}^{16}\times 0.0167\times 1\times {10}^{-9}\text{ J}\cdot {\text{s}}^{-1}{\text{m}}^{-2}\end{array}$

On simplifying the above equation,

$\text{1 W}=\text{1 J}\cdot {\text{s}}^{-1}$

$d\xi =1.01\times {10}^{-10}{\text{W/m}}^{\text{2}}$

Thus, the value of power flux is $1.18\times {10}^6{\text{W/m}}^{\text{2}}$.

Conclusion

The power of light at the temperatures $1000\text{K}$, $3000\text{K}$ and $10,000\text{K}$ is $1.01\times {10}^{-10}{\text{W/m}}^{\text{2}}$, $80.21{\text{W/m}}^{\text{2}}$ and $1.18\times {10}^6{\text{W/m}}^{\text{2}}$ respectively.