I have some contradicting sources for what exactly the ratio $\frac{A_{ji}}{B_{ji}}$, where $A_{ji}$ is the Einstein coefficient for spontaneous emission from upper state $j$ to lower state $i$, and $B_{ji}$ is the one for stimulated emission. From Laser Chemistry (2007, Telle et al, page 23), I have the following three expressions

$$ B_{ji} = B_{ij} \frac{g_i}{g_j} \\ B_{ij} = \frac{8 \pi^3 R^2}{3hg_i} \\ A_{ji} = \frac{64 \pi ^4 R^2}{3h \lambda^3g_j} $$

Using these to solve for the ratio $\frac{A_{ji}}{B_{ji}}$ results in

$$ \left(\frac{A_{ji}}{B_{ji}}\right)_{Derivation} = \frac{8 \pi \nu^3}{c^3} $$

where $\nu$ is the frequency and $c$ is the speed of light. However, Wikipedia disagrees on this expression, and says that

$$ \left(\frac{A_{ji}}{B_{ji}} \right)_{Wikipedia} = \frac{8 \pi h \nu^3}{c^3} $$

Further, my lecture notes have that

$$ \left(\frac{A_{ji}}{B_{ji}} \right)_{Lecture} = \frac{8 \pi h \nu}{c^3} $$

I am inclined to believe that my lecture notes have a typo, and should be the same as the Wikipedia source, and that my derivation is incorrect. Can I not use the expressions I did? Where does the Planck's constant come in? Is there a typo in the book? Should not this ratio be unitless?

Here (page 21) is another source that agrees with Wikipedia

$$ \left(\frac{A_{ji}}{B_{ji}} \right)_{UniversityofMaryland} = \frac{8 \pi h \nu^3}{c^3} $$


1 Answer 1


It cannot be unitless since the probabilty per unit time $B_{21}$ depends on the applied electromagnteic field energy density of frequency $\nu$ ie,$\rho(\nu)$. It is multipied with the energy density to make it the same dimensions as $A_{21}$..the probability per unit time of spontaneous emission. For thermal equilibrium, $n_1B_{12}\rho(\nu)=n_2[A_{21} + B_{21}\rho(\nu)]$. Solve it you get, $\rho(\nu)$=$\frac{\frac{A_{21}}{B_{21}}}{(n_1/n_2)*(B_{12}/B_{21}) - 1}$ which is equal to $\frac{8\pi h\nu^3}{c^3}$*$\frac{1}{e^\frac{h\nu}{KT}-1}$(from planck's radiation law) Compare the numerator and denominator for both the expressions for $\rho(\nu)$, you will see that $\frac{A_{21}}{B_{21}}$=$\frac{8\pi h\nu^3}{c^3}$ and, $\frac{n_1}{n_2}$ is given by the boltzmann factor $e^\frac{E_2-E_1}{KT}$ = $e^\frac{h\nu}{KT}$

  • $\begingroup$ So the probability of stimulated emission is much higher for lower wavelengths? Since $\frac{A_{ji}}{B_{ji}} \propto \frac{1}{\lambda^3}$ $\endgroup$
    – Yoda
    Commented Oct 1, 2016 at 13:30
  • $\begingroup$ You got that inverted. the probability of spontaneous emission is much higher for lower wavelengths since $A_{ji}$ is the probability for spontaneous emission; think of it this way, the higher the energy of the state the electron is in, the more the probability of it to jump to a lower energy state SPONTANEOUSLY $\endgroup$ Commented Oct 1, 2016 at 13:35
  • $\begingroup$ Yes, of course. I confused myself there! $\endgroup$
    – Yoda
    Commented Oct 1, 2016 at 13:42
  • $\begingroup$ if there isnt anymore confusion or doubts, could you mark it as done? $\endgroup$ Commented Oct 1, 2016 at 13:43

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