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

(N/A) $(i)$ Given,
Wavelength associated with the first transition,$\lambda_{1} = 589 \, nm = 589 \times 10^{-9} \, m$
Wavelength associated with the second transition,$\lambda_{2} = 589.6 \, nm = 589.6 \times 10^{-9} \, m$
Frequency of the first wavelength is $\nu_{1} = \frac{c}{\lambda_{1}} = \frac{3 \times 10^{8} \, m s^{-1}}{589 \times 10^{-9} \, m} = 5.093 \times 10^{14} \, s^{-1}$
Frequency of the second wavelength is $\nu_{2} = \frac{c}{\lambda_{2}} = \frac{3 \times 10^{8} \, m s^{-1}}{589.6 \times 10^{-9} \, m} = 5.088 \times 10^{14} \, s^{-1}$
$(ii)$ Energy difference between the two excited states is given as:
$\Delta E = h(\nu_{1} - \nu_{2})$
$\Delta E = 6.626 \times 10^{-34} \, J s \times (5.093 \times 10^{14} - 5.088 \times 10^{14}) \, s^{-1}$
$\Delta E = 6.626 \times 10^{-34} \, J s \times 0.005 \times 10^{14} \, s^{-1}$
$\Delta E = 3.313 \times 10^{-22} \, J$