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प्रश्न
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.
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उत्तर
For λ1 = 589 nm
Frequency of transition `("v"_1) = "c"/lambda_1`
`= (3.0xx10^(8) " ms"^(-1))/(589xx10^(-9) " m")`
Frequency of transition (ν1) = 5.093 × 1014 s–1
Similarly, for λ2 = 589.6 nm
Frequency of transition `("v"_2) = "c"/lambda_2`
`= (3.0 xx 10^8 " ms"^(-1))/(589.6xx10^(-9) " m")`
Frequency of transition (ν2) = 5.088 × 1014 s–1
Energy difference (ΔE) between excited states = E1 – E2
Where,
E2 = energy associated with λ2
E1 = energy associated with λ1
ΔE = hν1 – hν2
= h(ν1 – ν2)
= (6.626 × 10-34 Js) (5.093 × 1014 – 5.088 × 1014)s-1
= (6.626 × 10-34 J) (5.0 × 10-3 × 1014)
ΔE = 3.31 × 10-22 J
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