Einstein Coefficients

Einstein Coefficients:

Consider an enclosure containing atoms capable of being raised by absorption of radiation from level 'm' to excited level 'n'. In the absence of any radiation of appropriate frequency, there is no way in which atoms can pass from state 'm' to state 'n'.

Suppose now that the enclosure contains the isotropic radiation of frequency between $\omega_{mn}$ and $\omega_{mn}+d\omega_{mn}$ corresponding to the transition of atoms from state 'm' to state 'n' $(\hbar \omega_{mn}=E_n-E_m)$where $E_n$ is energy of state 'n' and $E_m$ is energy of state 'm'. Radiation as it is streaming out of the walls of the enclosure and back into them may interact with atoms and a transition from state 'm' to state 'n' may occur by absorption of a photon having energy $(\hbar \omega_{nm})$The rate of transition from state 'm' to 'n', $R_{mn}$ will be proportional to population of lower level per unit volume $(N_{m})$to probability of absorption per unit time $(B_{mn})$and to the density of radiation $\rho(\omega_{nm})$$.$That is,

$R_{mn}=\rho(\omega_{nm})B_{mn} N_m........................(1)$

$R_{mn}=\rho(\omega_{nm})B_{mn}N_m$

The atoms which absorb radiation become excited and eventually return to ground state emitting spontaneously their excess energy. The process is independent of radiation field energy. There is a finite probability $(A_{nm})$per unit time that the atom will spontaneously fall to the ground state emitting a photon $(\hbar \omega_{nm})$The number of atoms spontaneously emitting radiation per unit time is 

$R_{nm(sp)}=A_{nm} N_n........................(2)$
where $N_n$ is number of atoms per unit volume in state 'n'.

When a steady state reached, there must clearly be a balance achieved between absorption and emission processes. When the light and atoms have settled down in a steady equilibrium state, the mean number of atoms $N_m,N_n$ in lower ad upper states remains constant. That is, $\dfrac{dN}{dt}=0$

The number of atoms absorbing radiation per unit time= The number of atoms emitting radiation per unit time

If absorption and spontaneous emission were the only processes operative,

$\rho(\omega_{nm})B_{mn}N_m=A_{nm}N_n$

$\rho(\omega_{nm})=\dfrac{A_{nm}}{B_{mn}}.\dfrac{N_n}{N_m}..........................(3)$

In thermal equilibrium, the total number of atoms 'N' will be distributed over the available levels according to Boltzmann distribution as

$N_i=\dfrac{Ng_ie^{-E_i/kT}}{\sum_{i}g_ie^{-E_i/kT}}$

where $g_i$ is statistical wt. of level 'i'

$T$is temperature and $k$is Boltzmann constant

Therefore,

$\dfrac{N_n}{N_m}=\dfrac{g_ne^{-E_n/kT}}{g_me^{-E_m/kT}}$

Let's assume $g_n=g_m=1$

The above equation becomes,

$\dfrac{N_n}{N_m}=e^{-(E_n-E_m)/kT}$

$\dfrac{N_n}{N_m}=e^{-\hbar\omega_{nm}/kT}$

Put this value into the equation (3),

$\rho(\omega_{nm})=\dfrac{A_{nm}}{B_{mn}}e^{-\hbar\omega_{nm}/kT}.........................(4)$

Above equation for variation of radiation density $\rho(\omega_{nm})$with temperature is not in accordance with experimental measurements which support Planck's law which is

$\rho(\omega_{nm})d\omega_{nm}=\dfrac{\hbar\omega_{nm}^3}{\pi^2c^3}\dfrac{1}{e^{-\hbar \omega_{nm}/kT}}.d\omega_{nm}............................(5)$

Einstein's treatment of this problem uses an elegant thermodynamic argument. he pointed out that the two kinds of radiations indicates above are not sufficient for explaining the existence of state of equilibrium between radiation and matter.

The probability of spontaneous transition is determined only by internal properties of atoms, whereas that of absorption transition depends both on the prop. of atoms and on the density of radiation.

Therefore, while considering the detailed balance between two energy levels 'm' and 'n', in the presence of radiation, we can not ignore the possibility of radiation influencing atoms in excited state and causing them to return to lower state, the excess energy being given out as radiation. This radiative process is called induced or stimulated emission of radiation. Therefore,

$R_{nm(st)}=\rho(\omega_{nm})B_{nm}N_n........................(6)$

where $B_{nm}$ is the prob. per unit time that atom will undergo transition from 'n' to 'm' by stimulated emission. That is, absorption and stimulated emission are the processes induced by radiation.

The equation of rate of change of $N_n$ is

$\dfrac{dN_n}{dt}=-N_nA_{nm}+N_mB_{mn}\rho(\omega_{nm})-N_nB_{nm}\rho(\omega_{nm})$

Under the condition of equilibrium $\dfrac{dN_n}{dt}=0$

Therefore,

$N_nA_{nm}=(N_mB_{mn}-N_nB_{nm})\rho(\omega_{nm})$

$\rho(\omega_{nm})=\dfrac{A_{nm}}{\dfrac{N_m}{N_n}(B_{mn}-B_{nm})}...............(7)$

The ratio of number densitites of particles in two energy states in presence of degeneracy as obtained from Boltzmann law as

$\dfrac{N_n}{N_m}=\dfrac{g_ne^{-E_n/kT}}{g_me^{-E_m/kt}}=\dfrac{g_n}{g_m}e^{-(E_n-E_m)/kT}$

Put this value in equation (7)

$\rho(\omega_{nm})=\dfrac{A_{nm}}{\dfrac{g_m}{g_n}e^{-\hbar \omega/kT(B_{mn}-B_{nm})}}$

$\rho(\omega_{nm})=\dfrac{A_{nm}}{B_{nm}}\dfrac{1}{\dfrac{g_m}{g_n}\dfrac{B_{mn}}{B_{nm}}e^{(\hbar \omega/kT)}-1}.........................(8)$

This expression; derived using Einstein coefficient $A_{nm}, B_{nm}, B_{mn}$ and the Boltzmann distribution law must be consisted with Planck's law.

To determine $\dfrac{A_{nm}}{B_{nm}}$, Einstein took advantage of the fact that at high frequency eq. (8) must transform into Wein's radiation law. 

When $\hbar \omega \gt\gt kT$

$\rho(\omega_{nm})=\dfrac{\hbar \omega^3_{nm}}{\pi^2c^3}e^{-\hbar \omega_{nm}/kT}$

Thus, the equation (8) changes

$\rho(\omega_{nm})=\dfrac{A_{nm}}{B_{nm}}\dfrac{1}{\dfrac{g_m}{g_n}\dfrac{B_{mn}}{B_{nm}}e^{\hbar \omega/kT}}$

If we assume $g_mB_{mn}=g_nB_{nm}...........................(9)$

$\rho(\omega_{nm})\approx \dfrac{A_{nm}}{B_{nm}}e^{-\hbar \omega/kT}.............................(10)$

On comparing,

$\dfrac{A_{nm}}{B_{nm}}=\dfrac{\hbar \omega^3_{nm}}{\pi^2c^3}.........................(11)$

Equations (9),(11) make equation (8) identical with equation (5) and also shows three Einstein coefficient are interrelated. For non-degenerate states $g_n=g_m=1$

Therefore, $B_{nm}=B_{mn}$

That is, when an atom or a molecule with two energy levels is placed in electromagnetic field then prob. of upward and downward transitions are the same.