Calculate the ratio of the sodium atoms:

The features yellow emission of sodium vapours consists of a pair of lines at 589 nm and 589.6 nm. Those arise from the emission of radiation through the gaseous sodium atoms in the 3p excited state to 3s ground state. Calculate the ratio of the sodium atoms in the excited state to the ground state.

Solution

As per Eq. we have,

N*/N₀ =g*/g₀e^-ΔE/kT

Thus, to evaluate the ratio of the atoms within the excited state to that in the ground state we required to know the statistical factors g* and g₀ and ΔE.

As the 3s and 3p levels have two and six quantum states i.e., the statistical factors are 2 and 6 respectively, the ratio of g* and g0 comes out to be 6/2 = 3

The change in energy,  ΔE, can be calculated by using the formula,  ΔE = hc/ λ

ΔE =6.626 ×10 -34 Js × 3.0 ×108 ms ^-1/ 589.3×10 ^-9 m

The wavelength is taken as an average of the two, i.e., (589.0 + 589.6/2)

ΔE = 3.37 × 10 ^-19  J

Substituting the value of ΔE in the above equation, we get the following.

N*/N₀ =6/2e^{-3.37 ×10-19 J/1.38×10- 23  JK    × 3000 K}

N*/N₀ =3e^{-3.37 ×10-19/4.14 ×10- 20}

=   3e^-8.08 = 3.096 × 10^-4

You have learnt above that the intensity of a signal depends on the population of the level from which the transition originates and the probability of such a transition. These are the intrinsic parameters of the analyte being determined. Further to those, the intensity of a signal does depend on an important external parameter, viz., the concentration of the analyte. This, actually is the basis of quantitative aspect of the spectroscopic methods. Now let us see how the intensity depends on the concentration of the analyte for an emission spectrum.