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

For λ₁ = 589 nm

Frequency of transition

Frequency of transition (ν₁) = 5.093 × 10¹⁴ s^–1

Similarly, for λ₂ = 589.6 nm

Frequency of transition

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