Question

Use the data in Figure 4.8 to estimate the ratio of radiation intensity at $\mathbf{10}\mathbf{,}\mathbf{000}{\mathbf{\text{\hspace{0.17em}A}}}^{\mathbf{\circ }}$(infrared) to that at $\mathbf{5000}{\mathbf{\text{\hspace{0.17em}A}}}^{\mathbf{\circ }}$ (visible) from a blackbody at 5000 K. How will this ratio change with increasing temperature? Explain how this changeoccurs.

Short answer

The ratio of radiation intensity is 2.35.

When temperature increases, the ratio decreases.

Step-by-step solution

Blackbody radiation

A black body is an object where all radiations are absorbed in all wavelengths. The emission of the black body is termed black body radiation.

Step 2: Calculation of two frequencies

The radiant energy per unit volume in a black body is a function of temperature and frequency. It can be written as follows:

$\mathrm{\rho T}\left(\mathrm{\nu }\right)=\frac{8\mathrm{\pi }\text{h}{\mathrm{\nu }}^3}{{\text{c}}^3}\frac{1}{{\mathrm{e}}^{\frac{\text{h}\mathrm{\nu }}{{\text{k}}_{\text{B}}\text{T}}}-1}\Rightarrow \left(1\right)$

Its SI unit is ${\text{Jm}}^{-3}{\text{Hz}}^{-1}$

Now, we can calculate the two frequencies from wavelengths and velocity of light, given the first wavelength ${\lambda }_1=10,000{\text{\hspace{0.17em}A}}^{\circ }=10,000\times {10}^{-10}\text{\hspace{0.17em}m}$and second wavelength ${\lambda }_1=5000\times {10}^{-10}\text{\hspace{0.17em}m}$.

First frequency,

$\begin{array}{c}{\mathrm{\nu }}_1=\frac{\text{c}}{{\mathrm{\lambda }}_1}\\ =\frac{3\times {10}^8{\text{\hspace{0.17em}ms}}^{-1}}{10,000\times {10}^{-10}\text{\hspace{0.17em}m}}=3\times {10}^{14}{\text{\hspace{0.17em}s}}^{-1}\end{array}$

It is in the IR region.

Second frequency,

$\begin{array}{c}{\mathrm{\nu }}_2=\frac{\text{c}}{{\mathrm{\lambda }}_2}\\ =\frac{3\times {10}^8{\text{\hspace{0.17em}ms}}^{-1}}{5000\times {10}^{-10}\text{\hspace{0.17em}m}}=6\times {10}^{14}{\text{\hspace{0.17em}s}}^{-1}\end{array}$

It is in the visible region.

Calculation of ratio of radiation intensity

By dividing the equation (1) using ${\nu }_1$ by using ${\nu }_2$, we get

$\frac{\mathrm{\rho T}\left({\mathrm{\nu }}_1\right)}{\mathrm{\rho T}\left({\mathrm{\nu }}_2\right)}={\left(\frac{{\mathrm{\nu }}_1}{{\mathrm{\nu }}_2}\right)}^3\frac{{\mathrm{e}}^{\frac{\text{h}{\mathrm{\nu }}_2}{{\text{k}}_{\text{B}}\text{T}}}-1}{{\mathrm{e}}^{\frac{\text{h}{\mathrm{\nu }}_1}{{\text{k}}_{\text{B}}\text{T}}}-1}$

The value $\frac{\text{h}}{{\text{k}}_{\text{B}}}=4.80\times {10}^{-11}\text{\hspace{0.17em}Ks}$ . Substituting this value and temperature 5000 K in the above equation, we get

$\begin{array}{c}\frac{{\mathrm{\rho }}_{5000}\left({\mathrm{\nu }}_1\right)}{{\mathrm{\rho }}_{5000}\left({\mathrm{\nu }}_2\right)}={\left(\frac{3.0\times {10}^{14}{\text{\hspace{0.17em}s}}^{-1}}{6.0\times {10}^{14}{\text{\hspace{0.17em}s}}^{-1}}\right)}^3\frac{{\mathrm{e}}^{\frac{(\text{4.80}\times {\text{10}}^{-11}\text{\hspace{0.17em}Ks})(6.0\times {10}^{14}{\text{\hspace{0.17em}s}}^{-1})}{5000\text{\hspace{0.17em}K}}}-1}{{\mathrm{e}}^{\frac{(\text{4.80}\times {\text{10}}^{-11}\text{\hspace{0.17em}Ks})(3.0\times {10}^{14}{\text{\hspace{0.17em}s}}^{-1})}{5000\text{\hspace{0.17em}K}}}-1}\\ =\frac{1}{8}\left(\frac{{\mathrm{e}}^{5.76}-1}{{\mathrm{e}}^{2.88}-1}\right)\\ =\frac{1}{8}\left(\frac{317.35-1}{17.81-1}\right)=2.35\end{array}$

Calculation of ratio of radiation intensity with higher temperature

Now, we can use a higher temperature, 10,000 K, to find the ratio of radiation intensity.

$\begin{array}{c}\frac{{\mathrm{\rho }}_{10000}\left({\mathrm{\nu }}_1\right)}{{\mathrm{\rho }}_{10000}\left({\mathrm{\nu }}_2\right)}={\left(\frac{3.0\times {10}^{14}{\text{\hspace{0.17em}s}}^{-1}}{6.0\times {10}^{14}{\text{\hspace{0.17em}s}}^{-1}}\right)}^3\frac{{\mathrm{e}}^{\frac{(\text{4.80}\times {\text{10}}^{-11}\text{\hspace{0.17em}Ks})(6.0\times {10}^{14}{\text{\hspace{0.17em}s}}^{-1})}{10000\text{\hspace{0.17em}K}}}-1}{{\mathrm{e}}^{\frac{(\text{4.80}\times {\text{10}}^{-11}\text{\hspace{0.17em}Ks})(3.0\times {10}^{14}{\text{\hspace{0.17em}s}}^{-1})}{10000\text{\hspace{0.17em}K}}}-1}\\ =\frac{1}{8}\left(\frac{{\mathrm{e}}^{2.88}-1}{{\mathrm{e}}^{1.44}-1}\right)\\ =\frac{1}{8}\left(\frac{17.81-1}{4.22-1}\right)=0.652\end{array}$

It is obvious that when temperature increases, the ratio decreases. The exponentials also decrease, in which the numerator decreases faster. When temperature increases, the intensity shifts towards the blue region of the spectrum.