Chapter 9: Q8P (page 356)

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 $5 \times 10^{12} Hz$ , whereas spontaneous emission dominates for frequencies well above . Which mechanism dominates for visible light?

Short Answer

Expert verified

The spontaneous emission dominates.

Step by step solution

Formula for the spontaneous emission rate

The spontaneous emission rate is given by:

$A = \frac{ω^{3} {|℘|}^{2}}{3 {πo}_{0} h c^{3}}$

The thermally stimulated emission rate is given by:

role="math" localid="1658378927515" $R = \frac{π}{3 o_{0} h^{2}} {|℘|}^{2} ρ (ω)$

where:

$ρ (ω) = \frac{h}{π^{2} c^{3}} \frac{ω^{3}}{(e^{h ω / k_{B} T} - 1)}$

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

$\frac{A}{R} = \frac{ω^{3} {|℘|}^{2}}{3 {πo}_{0} h c^{3}} . \frac{3 {πo}_{0} h^{2}}{π {|℘|}^{2}} . \frac{π^{2} c^{3} (e^{h ω / K_{B} T} - 1)}{h ω^{3}} = e^{h ω / K_{B} T} - 1$

Find out the result

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

$1 = e^{h ω / K_{B} T} - 1 e^{h ω / K_{B} T} - 2 \frac{h ω}{k_{B} T} = I n 2 ω = \frac{k_{B} T}{h} I n 2$

The frequency is,

$v = \frac{ω}{2 π} = \frac{k_{B} T}{h} I n 2$

At room temperature, we get:

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

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

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