Derivation of Einstein Coefficient Relation→
Let us consider the $N_{1}$ and $N_{2}$ is the mean population of lower energy state and upper energy state respectively. If the energy density of incident light is $\rho(\nu)$ then
The rate of transition of number of atoms due to absorption process:
$R_{abs}=B_{12} \: \rho(v) \: N_{1} \qquad(1)$
The above equation shows the number of atoms absorbing the photon per second per unit volume
Where $B_{12}$= Einstein Absorption Coefficent
The rate of transition of number of atoms due to sponteneous emission process:
$R_{sp}=A_{21} \: N_{2} \qquad(2)$
The above equation shows the number of atoms emitting the photon per second per unit volume due to spontaneous emission
Where $A_{21}$= Einstein Spontaneous Emission Coefficient
The rate of transition of the number of atoms due to stimulated emission process:
$R_{st}=B_{21} \: \rho(v) \: N_{2} \qquad(3)$
The above equation shows the number of atoms emitting the photon per second per unit volume due to stimulated emission
Where $B_{21}$= Einstein Stimulated Emission Coefficient
Under the thermal equilibrium, the mean population $N_{1}$ and $N_{2}$ in lower and upper energy states respectively must remain constant. This condition requires that the transition of the number of atoms from $E_{2}$ to $E_{1}$ must be equal to the transition of the number of atoms from $E_{1}$ to $E_{2}$. Thus
$\left.\begin{matrix}The \: number \: of \: atoms \: absorbing \\ photons \: per \: second \: per \: unit \: volume
\end{matrix}\right\} = \left.\begin{matrix} The \: number \: of \: atoms \: emitting \\ photons \: per \: second \: per \: unit \: volume
\end{matrix}\right\}$
i.e $R_{abs}= R_{sp}+R_{st}$
$B_{12} \: \rho(v) \: N_{1}= A_{21} \: N_{2} + B_{21} \: \rho(v) \: N_{2}$
$B_{12} \: \rho(v) \: N_{1}  B_{21} \: \rho(v) \: N_{2} = A_{21} \: N_{2} $
$ \rho(v) (B_{12} \: N_{1}  B_{21} \: N_{2} ) = A_{21} \: N_{2} $
$\rho(v)=\frac{A_{21} \: N_{2}}{(B_{12} \: N_{1}  B_{21} \: N_{2} )} \qquad(4)$
We know that
$\frac{N_{1}}{N_{2}}=e^{\frac{(E_{2}E_{1})}{kT}}$
$\frac{N_{1}}{N_{2}}=e^{\frac{h\nu}{kT}}$
Now substitute the value of $\frac{N_{1}}{N_{2}}$ in equation $(4)$
$\rho(v)=\frac{A_{21}}{B_{12}} \left [ \frac{1}{e^{\frac{h\nu}{kT}} \frac{B_{21}}{B_{12}}} \right ] \qquad(5)$
According to Planck's Radiation Law
$\rho(v)=\frac{8\pi h \nu^{3}}{c^{3}} \left [ \frac{1}{e^{\frac{h\nu}{kT}} 1} \right ] \qquad(6)$
Now comparing the equation $(5)$ and equation $(6)$
$\frac{B_{21}}{B_{12}}=1$ and $\frac{A_{21}}{B_{12}}=\frac{8\pi h \nu^{3}}{c^{3}}$
From the above equation, we get
$B_{12}=B_{21}=\frac{c^{3}}{8\pi h \nu^{3}}A_{21}$
