Question

As a mechanism for downward transitions, spontaneous emission competes with thermally stimulated emission (stimulated emission for which blackbody radiation is the source). Show that at room temperature (T = 300 K) thermal stimulation dominates for frequencies well below $\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{12}}\mathbf{Hz}$ , whereas spontaneous emission dominates for frequencies well above . Which mechanism dominates for visible light?

Short answer

The spontaneous emission dominates.

Step-by-step solution

Formula for the spontaneous emission rate

The spontaneous emission rate is given by:

$A=\frac{{\omega }^3{\left|\wp \right|}^2}{3{\mathrm{\pi o}}_0\sout{\mathrm{h}}{\mathrm{c}}^3}$

The thermally stimulated emission rate is given by:

role="math" localid="1658378927515" $R=\frac{\mathrm{\pi }}{3o_0\sout{h^2}}{\left|\wp \right|}^2\rho \left(\omega \right)$

where:

$\rho \left(\omega \right)=\frac{\sout{h}}{{\mathrm{\pi }}^2{\mathrm{c}}^3}\frac{{\omega }^3}{\left(e^{\sout{h}\omega /k_{B}T}-1\right)}$

The ratio of the spontaneous emission rate to the thermally stimulated emission rate is,

$$\begin{gathered} \frac{A}{R}=\frac{{\omega }^3{\left|\wp \right|}^2}{3{\mathrm{\pi o}}_0\sout{\mathrm{h}}{\mathrm{c}}^3}.\frac{3{\mathrm{\pi o}}_0{\sout{\mathrm{h}}}^2}{\mathrm{\pi }{\left|\mathrm{\wp }\right|}^2}.\frac{{\mathrm{\pi }}^2{\mathrm{c}}^3\left({\mathrm{e}}^{\sout{\mathrm{h}}\mathrm{\omega }/{\mathrm{K}}_{\mathrm{B}}\mathrm{T}}-1\right)}{\sout{h}{\omega }^3} \\ ={\mathrm{e}}^{\sout{\mathrm{h}}\mathrm{\omega }/{\mathrm{K}}_{\mathrm{B}}\mathrm{T}}-1 \end{gathered}$$

Find out the result

Seek the point at which the ratio is, that is:

$$\begin{gathered} 1={\mathrm{e}}^{\sout{\mathrm{h}}\mathrm{\omega }/{\mathrm{K}}_{\mathrm{B}}\mathrm{T}}-1 \\ {\mathrm{e}}^{\sout{\mathrm{h}}\mathrm{\omega }/{\mathrm{K}}_{\mathrm{B}}\mathrm{T}}-2 \\ \frac{\sout{h}\omega }{k_{B}T}=In2 \\ \omega =\frac{k_{B}T}{\sout{h}}In2 \end{gathered}$$

The frequency is,

$$\begin{gathered} v=\frac{\omega }{2\mathrm{\pi }} \\ =\frac{k_{B}T}{h}In2 \end{gathered}$$

At room temperature, we get:

$$\begin{gathered} \mathrm{v}=\frac{\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{k}\right)\left(300\mathrm{k}\right)}{6.63\times {10}^{-34}\mathrm{J}.\mathrm{s}} \\ =4.33\times {10}^{12}\mathrm{Hz} \end{gathered}$$

For a frequency higher than this frequency the spontaneous emission dominates, note that higher frequencies than this one includes the visible light.