Chapter 40, Problem 11P

To determine

The number of photons per second escaping the opening and having wavelengths between $500\text{ nm}$ and $501\text{ nm}$ .

Answer to Problem 11P

The number of photons per second escaping the opening and having wavelengths between $500\text{ nm}$ and $501\text{ nm}$  is $1.30\times {10}^{15}\text{ photons/s}$ .

Explanation of Solution

The wavelength range is between $500\text{ nm}$ and $501\text{ nm}$ . Since this range of wavelength is small, the wavelength in the Plank’s law can be taken as the average of these two wavelengths.

Write the Plank’s radiation law.

  $I\left(\overline{\lambda },T\right)=\frac{2\pi h{c}^2}{{\overline{\lambda }}^5\left(e^{hc/\overline{\lambda }{k}_{B}T}-1\right)}$                                                                                          (I)

Here, $I\left(\overline{\lambda },T\right)$ is the intensity per wavelength, $h$ is the Planck’s constant, $c$ is the speed of the photon, $\overline{\lambda }$ is the average wavelength of the radiation emitted, $k_B$ is the Boltzmann constant and $T$ is the temperature of the blackbody.

Write the equation for $\overline{\lambda }$ .

  $\overline{\lambda }=\frac{{\lambda }_1+{\lambda }_2}{2}$                                                                                                             (II)

Here, ${\lambda }_1$ is the lower value of the wavelength and ${\lambda }_2$ is the higher value of the wavelength.

Put equation (II) in equation (I).

  $I\left(\overline{\lambda },T\right)=\frac{2\pi h{c}^2}{{\left[\left({\lambda }_1+{\lambda }_2\right)/2\right]}^5\left(e^{hc/\left[\left({\lambda }_1+{\lambda }_2\right)/2\right]k_{B}T}-1\right)}$                                                           (III)

The energy per time leaving the hole can be determined by taking the product of the intensity per wavelength, area and the wavelength range.

Write the equation for the energy per time leaving the hole.

  $P=I\left(\overline{\lambda },T\right)\left({\lambda }_2-{\lambda }_1\right)A$                                                                                           (IV)

Here, $P$ is the energy per time leaving the hole and $A$ is the area of the hole.

Write the equation for the area of the hole.

  $\begin{array}{c}A=\pi {\left(\frac{d}{2}\right)}^2\\ =\pi \frac{d^2}{4}\end{array}$                                                                                                           (V)

Here, $d$ is the diameter of the hole.

Put equations (III) and (V) in equation (IV).

  $\begin{array}{c}P=\frac{2\pi h{c}^2}{{\left[\left({\lambda }_1+{\lambda }_2\right)/2\right]}^5\left(e^{hc/\left[\left({\lambda }_1+{\lambda }_2\right)/2\right]k_{B}T}-1\right)}\left({\lambda }_2-{\lambda }_1\right)\pi \frac{d^2}{4}\\ =\frac{16{\pi }^{2}h{c}^2}{{\left({\lambda }_1+{\lambda }_2\right)}^5\left(e^{2hc/\left({\lambda }_1+{\lambda }_2\right)k_{B}T}-1\right)}\left({\lambda }_2-{\lambda }_1\right)d^2\end{array}$                                              (VI)

Write the equation for the average photon energy.

  $\overrightarrow{E}=h\overrightarrow{f}$                                                                                                                   (VII)

Here, $\overline{E}$ is the average photon energy and $\overline{f}$ is the average photon frequency.

Write the equation for $\overline{f}$ .

  $\overline{f}=\frac{c}{\overline{\lambda }}$

Put equation (II) in the above equation.

  $\begin{array}{c}\overline{f}=\frac{c}{\left({\lambda }_1+{\lambda }_2\right)/2}\\ =\frac{2c}{{\lambda }_1+{\lambda }_2}\end{array}$

Put the above equation in equation (VII).

  $\begin{array}{c}\overrightarrow{E}=h\left(\frac{2c}{{\lambda }_1+{\lambda }_2}\right)\\ =\frac{2hc}{{\lambda }_1+{\lambda }_2}\end{array}$                                                                                                    (VIII)

Write the equation for the number of photons per second escaping the opening.

  $n=\frac{P}{\overline{E}}$

Here, $n$ is the number of photons per second escaping the opening.

Put equations (VI) and (VIII) in the above equation.

  $\begin{array}{c}n=\frac{\frac{16{\pi }^{2}h{c}^2}{{\left({\lambda }_1+{\lambda }_2\right)}^5\left(e^{2hc/\left({\lambda }_1+{\lambda }_2\right)k_{B}T}-1\right)}\left({\lambda }_2-{\lambda }_1\right)d^2}{\frac{2hc}{{\lambda }_1+{\lambda }_2}}\\ =\frac{8{\pi }^{2}c{d}^2\left({\lambda }_2-{\lambda }_1\right)}{{\left({\lambda }_1+{\lambda }_2\right)}^4\left(e^{2hc/\left({\lambda }_1+{\lambda }_2\right)k_{B}T}-1\right)}\end{array}$                                                           (IX)

Conclusion:

It is given that the temperature of the blackbody is $7500\text{ K}$ and the diameter of the opening is $0.0500\text{ mm}$ . The value of $h$ is $6.626\times {10}^{-34}\text{ J}\cdot \text{s}$ , the speed of photon is $3.00\times {10}^8\text{ m/s}$ and the value of $k_B$ is $1.38\times {10}^{-23}\text{ J/K}$ .

Substitute  , $3.00\times {10}^8\text{ m/s}$ for $c$ , $0.0500\text{ mm}$ for $d$ , $501\text{ nm}$ for ${\lambda }_2$ , $500\text{ nm}$ for ${\lambda }_1$ , $6.626\times {10}^{-34}\text{ J}\cdot \text{s}$ for $h$ , $1.38\times {10}^{-23}\text{ J/K}$ for $k_B$ and $7500\text{ K}$ for $T$ in equation (IX) to find $n$ .

    $\begin{array}{c}n=\frac{8{\pi }^2\left(3.00\times {10}^8\text{ m/s}\right){\left(0.0500\text{ mm}\frac{1\text{ m}}{1000\text{ mm}}\right)}^2\left(501\text{ nm}\frac{1\text{ m}}{{10}^9\text{ nm}}-500\text{ nm}\frac{1\text{ m}}{{10}^9\text{ nm}}\right)}{{\left(500\text{ nm}\frac{1\text{ m}}{{10}^9\text{ nm}}+501\text{ nm}\frac{1\text{ m}}{{10}^9\text{ nm}}\right)}^4\left(e^{2\left(6.626\times {10}^{-34}\text{ J}\cdot \text{s}\right)\left(3.00\times {10}^8\text{ m/s}\right)/\left(500\text{ nm}\frac{1\text{ m}}{{10}^9\text{ nm}}+501\text{ nm}\frac{1\text{ m}}{{10}^9\text{ nm}}\right)\left(1.38\times {10}^{-23}\text{ J/K}\right)\left(7500\text{ K}\right)}-1\right)}\\ =1.30\times {10}^{15}\text{ photons/s}\end{array}$

Therefore, the number of photons per second escaping the opening and having wavelengths between $500\text{ nm}$ and $501\text{ nm}$  is $1.30\times {10}^{15}\text{ photons/s}$ .