Derive an expression for energy density at thermal equilibrium through Einstein's coefficients.
Answers
Answer:
The Einstein coefficients are fixed probabilities per time associated with each atom, and do not depend on the state of the gas of which the atoms are a part. Therefore, any relationship that we can derive between the coefficients at, say, thermodynamic equilibrium will be valid universally.
At thermodynamic equilibrium, we will have a simple balancing, in which the net change in the number of any excited atoms is zero, being balanced by loss and gain due to all processes. With respect to bound-bound transitions, we will have detailed balancing as well, which states that the net exchange between any two levels will be balanced. This is because the probabilities of transition cannot be affected by the presence or absence of other excited atoms. Detailed balance (valid only at equilibrium) requires that the change in time of the number of atoms in level 1 due to the above three processes be zero:
{\displaystyle 0=A_{21}n_{2}+B_{21}n_{2}\rho (\nu )-B_{12}n_{1}\rho (\nu ).}{\displaystyle 0=A_{21}n_{2}+B_{21}n_{2}\rho (\nu )-B_{12}n_{1}\rho (\nu ).}
Along with detailed balancing, at temperature T we may use our knowledge of the equilibrium energy distribution of the atoms, as stated in the Maxwell–Boltzmann distribution, and the equilibrium distribution of the photons, as stated in Planck's law of black body radiation to derive universal relationships between the Einstein coefficients.
From Boltzmann distribution we have for the number of excited atomic species i:
{\displaystyle {\frac {n_{i}}{n}}={\frac {g_{i}e^{-E_{i}/kT}}{Z}},}{\displaystyle {\frac {n_{i}}{n}}={\frac {g_{i}e^{-E_{i}/kT}}{Z}},}
where n is the total number density of the atomic species, excited and unexcited, k is Boltzmann's constant, T is the temperature, {\displaystyle g_{i}}g_{i} is the degeneracy (also called the multiplicity) of state i, and Z is the partition function. From Planck's law of black-body radiation at temperature T we have for the spectral radiance (radiance is energy per unit time per unit solid angle per unit projected area, when integrated over an appropriate spectral interval)[24] at frequency ν
{\displaystyle \rho _{\nu }(\nu ,T)=F(\nu ){\frac {1}{e^{h\nu /kT}-1}},}{\displaystyle \rho _{\nu }(\nu ,T)=F(\nu ){\frac {1}{e^{h\nu /kT}-1}},}
where[25]
{\displaystyle F(\nu )={\frac {2h\nu ^{3}}{c^{3}}},}{\displaystyle F(\nu )={\frac {2h\nu ^{3}}{c^{3}}},}
where {\displaystyle c}c is the speed of light and {\displaystyle h}h is Planck's constant.
Substituting these expressions into the equation of detailed balancing and remembering that E2 − E1 = hν yields
{\displaystyle A_{21}g_{2}e^{-h\nu /kT}+B_{21}g_{2}e^{-h\nu /kT}{\frac {F(\nu )}{e^{h\nu /kT}-1}}=B_{12}g_{1}{\frac {F(\nu )}{e^{h\nu /kT}-1}},}{\displaystyle A_{21}g_{2}e^{-h\nu /kT}+B_{21}g_{2}e^{-h\nu /kT}{\frac {F(\nu )}{e^{h\nu /kT}-1}}=B_{12}g_{1}{\frac {F(\nu )}{e^{h\nu /kT}-1}},}
separating to
{\displaystyle A_{21}g_{2}(e^{h\nu /kT}-1)+B_{21}g_{2}F(\nu )=B_{12}g_{1}e^{h\nu /kT}F(\nu ).}{\displaystyle A_{21}g_{2}(e^{h\nu /kT}-1)+B_{21}g_{2}F(\nu )=B_{12}g_{1}e^{h\nu /kT}F(\nu ).}
The above equation must hold at any temperature, so
{\displaystyle B_{21}g_{2}=B_{12}g_{1},}{\displaystyle B_{21}g_{2}=B_{12}g_{1},}
and
{\displaystyle -A_{21}g_{2}+B_{21}g_{2}F(\nu )=0.}{\displaystyle -A_{21}g_{2}+B_{21}g_{2}F(\nu )=0.}
Therefore, the three Einstein coefficients are interrelated by
{\displaystyle {\frac {A_{21}}{B_{21}}}=F(\nu )}{\displaystyle {\frac {A_{21}}{B_{21}}}=F(\nu )}
and
{\displaystyle {\frac {B_{21}}{B_{12}}}={\frac {g_{1}}{g_{2}}}.}{\displaystyle {\frac {B_{21}}{B_{12}}}={\frac {g_{1}}{g_{2}}}.}
When this relation is inserted into the original equation, one can also find a relation between {\displaystyle A_{21}}A_{21} and {\displaystyle B_{12}}B_{12}, involving Planck's law.
Explanation:
HOPE IT HELPS YOU!
MARK AS BRAINLINEST