Chapter 4, Problem 4.1P

To determine

The sketch of the graph for the given function and to explain the trends in the graph.

Answer to Problem 4.1P

The graph for the function $I\left(\lambda ,T\right)=\frac{2\pi h{c}^2}{{\lambda }^5}\frac{1}{e^{hc/\lambda k_{\text{B}}T}-1}$ is shown in figure (1).

Explanation of Solution

Given:

The Planck distribution function for blackbody radiation is given by,

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

Calculation:

Consider the term,

  $\frac{2\pi h{c}^2}{{\lambda }^5}$

The value of the term is calculated as,

  $\begin{array}{c}\frac{2\pi h{c}^2}{{\lambda }^5}=\frac{2\pi \left(6.63\times {10}^{-34}\text{J}-\text{s}\right){\left(3\times {10}^8\text{m}/\text{s}\right)}^2}{{\lambda }^5}\\ =\frac{3.75\times {10}^{-16}\text{J}-{\text{m}}^{\text{2}}/\text{s}}{{\lambda }^5}\end{array}$

Consider the term,

  $\frac{hc}{k_{\text{B}}T}$

The value of the term is calculated as,

  $\begin{array}{c}\frac{hc}{k_{\text{B}}T}=\frac{\left(6.63\times {10}^{-34}\text{J}-\text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(1.38\times {10}^{-23}\text{J}/\text{K}\right)\left(1500\text{K}\right)}\\ =9.6087\times {10}^{-6}\text{m}\end{array}$

Thus, the Planck's distribution function of a blackbody radiation for a fixed temperature $T=1500\text{K}$ is given by,

  $I\left(\lambda ,T\right)=\frac{3.75\times {10}^{-16}\text{J}-{\text{m}}^{\text{2}}/\text{s}}{{\lambda }^5}\frac{1}{e^{\left(9.6087\times {10}^{-6}\text{m}\right)/\lambda }-1}$

The following table shows the Planck's intensity distribution for different wavelengths at a fixed temperature.

Wavelength $\left(\lambda ,\text{m}\right)$	$I\left(\lambda \right)$
$0$	$0$
$0.00001$	$2323463782$
$0.00002$	$189999621$
$0.00005$	$\begin{array}{c}5663546.65\end{array}$
$0.0001$	$\begin{array}{c}371821.542\end{array}$
$0.0002$	$\begin{array}{c}23810.7114\end{array}$

The following figure shows the Planck's intensity distribution for different wavelengths at a fixed temperature.

  

Figure (1)

From the graph, it can be concluded that the intensity of blackbody radiation for a fixed temperature, is maximum at a particular wavelength and it becomes zero with the increase in wavelength. With the increase in temperature, the maximum intensity of a radiation shift towards the shorter wavelengths.

Conclusion:

Therefore, the graph for the function $I\left(\lambda ,T\right)=\frac{2\pi h{c}^2}{{\lambda }^5}\frac{1}{e^{hc/\lambda k_{\text{B}}T}-1}$ is shown in figure (1).