Q. 50

# 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 :

Given:

Wavelength, λ_{1} = 589 nm

Wavelength, λ_{2} = 589.6 nm

To find frequency:

Speed of Light = [Frequency] × [Wavelength]

We know speed of light = 3×10^{8} m/s

Frequency, v_{1} = [3×10^{8}] / [589 × 10^{-9}]

v_{1} = 5.093×10^{14} Hz

Frequency, v_{2} = [3×10^{8}] / [589.6 × 10^{-9}]

V_{2} = 5.088×10^{14} Hz

Change in Energy, ∆E = E_{2}-E_{1} = h[v_{2}-v_{1}]

= [6.626× 10^{-34}] {[5.093 - 5.088] × 10^{14}}

= [6.626× 10^{-34}] {[5× 10^{-3}] × 10^{14}}

= 3.313× 10^{-22}J

Therefore, the energy difference between two excited states is 3.31× 10^{-22}J.

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