91Ó°ÊÓ

Consider any two internal states, s₁ and s₂, of an atom. Let s₂ be the higher-energy state, so that $E\left(s_2\right)-E\left(s_1\right)=\mathrm{Ï\mu }$for some positive constant. If the atom is currently in state s₂, then there is a certain probability per unit time for it to spontaneously decay down to state s₁, emitting a photon with energy e. This probability per unit time is called the Einstein A coefficient:

(a) Consider N₁ and N₂ atoms in state s₁ and s₂, respectively. The probability of atoms decaying to state s1 equals the coefficient A, and the rate of atoms per unit time equals the number of atoms in this state N₂ multiplied by the constant A, resulting in:

$A{N}_2$

Second, when we shine a light with energy e, there is a chance that the atoms in state s₁ will absorb the photons and jump to state s₂. The probability of this action occurring equals the constant B multiplied by the energy density of the beam u(f), so the rate per unit time of the atoms jumping to the second state equals the negative sign of the probability multiplied by the number of the first state, that is:

$-N_{1}Bu\left(f\right)$

Finally, we have stimulated emission from state s₂ to state s₁, which has a probability of the constant B' multiplied by the energy density of the beam u(f), so the rate per unit time of atoms jumped to the second state equals the probability multiplied by the number of atoms in the first state, i.e.

$N_2{B}^{\text{'}}u\left(f\right)$

The total rate $d{N}_1/dt$equals to the sum of these rates:

role="math" localid="1647754521419" $\frac{d{N}_1}{dt}=A{N}_2+N_2{B}^{\text{'}}u\left(f\right)-N_{1}Bu\left(f\right)\left(1\right)$

(b) Because the number of atoms N1 and N2 in the equilibrium is independent of time, therefore

$$\begin{gathered} \frac{d{N}_1}{dt}=0 \\ A{N}_2+N_2{B}^{\text{'}}u\left(f\right)-N_{1}Bu\left(f\right)=0 \\ A{N}_2+\left(N_2{B}^{\text{'}}-N_{1}B\right)u\left(f\right)=0\left(2\right) \end{gathered}$$

the ratio of the number of atoms can be written in terms of Boltzmann factor, as:

$$\begin{gathered} \frac{N_2}{N_1}=\frac{e^{-E\left(s_2\right)/kT}}{e^{-E\left(s_1\right)/kT}} \\ \frac{N_2}{N_1}=e^{-\left(E\left(s_2\right)-E\left(s_1\right)\right)/kT} \end{gathered}$$

But the energy difference between the levels is $\mathrm{Ï\mu }=hf$, therefore

$\frac{N_2}{N_1}=e^{-hf/kT}\left(3\right)$

Equation 7.83 is :

$\frac{U}{V}=\frac{8\mathrm{Ï€}}{(hc{)}^3}{∫}_0^{∞}\frac{{\mathrm{Ï\mu }}^3}{e^{\mathrm{Ï\mu }/kT}-1}d\mathrm{Ï\mu }$

Let $\mathrm{Ï\mu }=hf→d\mathrm{Ï\mu }=hdf$, then

$$\begin{gathered} \frac{U}{V}=\frac{8\mathrm{π}}{(hc{)}^3}{∫}_0^{∞}\frac{(hf{)}^3}{e^{hf/kT}-1}hdf \\ \frac{U}{V}=\frac{8\mathrm{π}h}{c^3}{∫}_0^{∞}\frac{f^3}{e^{hf/kT}-1}df \\ u\left(f\right)=\frac{8\mathrm{π}h}{c^3}\frac{f^3}{e^{hf/kT}-1} \end{gathered}$$

Substitute this into (2)

$A{N}_2+\left(N_2{B}^{\text{'}}-N_{1}B\right)\frac{8\mathrm{Ï€}h}{c^3}\frac{f^3}{e^{hf/kT}-1}=0$

Divide both sides by N₁and substitute from (3)

$A{e}^{-hf/kT}+\left(B^{\text{'}}{e}^{-hf/kT}-B\right)\frac{8\mathrm{Ï€}h}{c^3}\frac{f^3}{e^{hf/kT}-1}=0$

Multiply both sides by $e^{hf/kT}$

$$\begin{gathered} A+\left(B^{\text{'}}-B{e}^{hf/kT}\right)\frac{8\mathrm{Ï€}h}{c^3}\frac{f^3}{e^{hf/kT}-1}=0 \\ A=\left(B{e}^{hf/kT}-B^{\text{'}}\right)\frac{8\mathrm{Ï€}h}{c^3}\frac{f^3}{e^{hf/kT}-1} \end{gathered}$$

The constant A is temperature independent, as are the constants B and B'. This indicates that the RHS must be temperature independent, thus the temperature portion must cancel out. This only happens when B = B', so we get:

$$\begin{gathered} A=B\left(e^{hf/kT}-1\right)\frac{8\mathrm{Ï€}h}{c^3}\frac{f^3}{e^{hf/kT}-1} \\ A=B\frac{8\mathrm{Ï€}h{f}^3}{c^3} \end{gathered}$$