Question

The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calculate the frequency of each transition and energy difference between two excited states.

Answer 1

Frequency of transition ($\nu_1$) = $\frac{c}{\lambda_1}$ = $\frac{3.0\times10^8ms^{-1}}{589\times10^{-9}m}$
Frequency of transition ($\nu_1$) = 5.093 × 1014 s-1
Frequency of transition ($\nu_2$) = $\frac{c}{\lambda_2}$ = $\frac{3.0\times10^8ms^{-1}}{589.6\times10^{-9}m}$
Frequency of transition ($\nu_2$) = 5.088 × 1014 s-1
Energy difference (ΔE) between excited states = E1-E2
Where, E2 = energy associated with $\nu_2$
E1 = energy associated with $\nu_1$
ΔE= E1-E2= 6.626×10−34×(5.093−5.088)×1014
ΔE = 3.31 × 10-22 J