THE INTENSITIES OF ROTATIONAL SPECTRAL LINES

The probability of all rotational spectra changes with ∆J = ± 1 is almost the same value. This doesn't mean that all the lines will be equally intense. Although the intrinsic probability is same for all transitions, there is differences in the transition probability. Hence the intensity of the lines will be directly to the over population which is governed by the Blotzmann-Distribution law.

NJ/ N₀ = e⁻ ᴱᴶ/ᴷᵀ

= e⁻ᴮʰᶜʲ ( J+1)/KT

Now, taking a typical value of,

B= 2cm⁻¹ at 298 kelvin. We can calculate the NJ/ N₀ as

NJ/N₀ = exp [− ( 2×6.626×10⁻³⁴ × 3×10¹⁰ × 1(1+1)/1.38×10⁻²³× 298

= 0.98

Thus the population of ground and excited States are almost the same.

                                       Another factor which has to be taken in to account is the degeneracy of the rotational levels. The population at the energy EJ will be,

Population α (2J+1) (exp- EJ/kT)

When the population is plotted against J, the following curve below is obtained that the population rises to a maximum and then decreases.

Using the principle of maxima and minima, we can find the levels for which the population is maximum. The differentiate population with respect J and equate the result to zero.

dN/dJ = (2J+1) (exp- BhcJ(J+1)/kT [(2J+1)

(−Bhc/kT)] + 2exp [−BhcJ(J+1)] = 0

                                      kT

exp(-BhcJ(J+1)/kT[(2J+1)² (−Bhc/kT)+2] =0

(2J+1)²Bhc/kT=2: (2J+1)² = (2kT/Bhc)

(2J+1) = (2kT/Bhc)¹/₂; 2J= (2kT/Bhc)¹/₂−1;

J = (kT/2Bhc)¹/₂ − 1/2

Knowing the values of B and T, now we can finding the J for which the population is maximum and it is found to be 9.

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