Question

For a system in thermal equilibrium use Boltzmann's relationship to calculate the ratio of the number of atoms in energy states E1 and E2 if the wavelength of light produced by a transition between the two is 633nm at 300K. If the temperature is reduced to 200K what is the new ratio? Explain the significance of the calculated ratios above in the production of laser light.

Answer 1

The energy associated with temperature 300 K = k T , where k = Boltzman Constant = 8.617 x 10^(-5) eV/K

= 8.617 x 10^(-5) eV/K x 300 K = 0.0259 eV

The energy associated with temperature 200 K = k T = 8.617 x 10^(-5) eV/K x 200 K = 0.0172 eV

The energy associated with 633 nM = $hC/\lambda$ = 4.1357 x 10^(-15) eV s x 3 x 10^8 m/s / [633 x 10^(-9)]

= 1.96 eV

If N2 is no. of atoms at higher Energy level E2 and N1 is no.of atoms in Lower energy level E2

$N_{2}/N_{1} = e^{- (E2-E1/kT)}$

$N_{2}/N_{1} = e^{- (1.96/0.0259)}$ for 300 K  

it is obvious that N2 is almost zero in above equation

Even at 200K the value of N2 is negligible quantity.

These ratios show that there is no way we can increase the population of atoms in Higher state high due to thermal excitation

We have to other means of exciting the atoms to higher energy levels with sustained life time of the higher energy state to produce population inversion.