The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Energy difference between two excited states is

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$3.31 \times 10^{- 22} J$

$3.0 \times 10^{- 21} kJ$

$3.31 \times 10^{- 22} kJ$

$2.98 \times 10^{- 21} J$

answer is B.

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Detailed Solution

step I : Wavelength ( $λ$ ) = 589 nm $= 589 \times 10^{- 9} m$
Frequency $(v) = \frac{c}{λ} = \frac{3 \times 10^{8}}{589 \times 10^{- 9}}$
$= 5 .093 \times 10^{14} cycles per sec$
step II : Wavelenght $(λ) = 589 .6 nm = 589 .6 \times 10^{- 9} m$
$∴ v = \frac{c}{λ} = \frac{3 \times 10^{8}}{589 .6 \times 10^{- 9}} = 5 .088 \times 10^{14}$ cycles per sec
Energy difference between two excited states,
$\begin{array}{l} ΔE = 6 .626 \times 10^{- 34} (5 .093 - 5 .088) 10^{14} \\ = 6 .626 \times 10^{- 34} \times 5 \times 10^{- 3} \times 10^{14} = 3 .31 \times 10^{- 22} J \end{array}$