Q.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.

Concept: Photon Energy Calculation — the energy of a photon is directly proportional to its frequency ($E = h\nu$) and inversely proportional to its wavelength ($\nu = c/\lambda$). The energy difference between two excited states equals the difference in photon energies of the two absorption lines.

Step 1: Convert wavelengths to meters.

$\lambda_1 = 589\ \text{nm} = 589 \times 10^{-9}\ \text{m}$, $\lambda_2 = 589.6\ \text{nm} = 589.6 \times 10^{-9}\ \text{m}$.

Step 2: Calculate frequencies using $\nu = c/\lambda$ with $c = 3.00 \times 10^8\ \text{m/s}$.

$\nu_1 = \frac{3.00 \times 10^8}{589 \times 10^{-9}} = 5.093 \times 10^{14}\ \text{Hz}$

$\nu_2 = \frac{3.00 \times 10^8}{589.6 \times 10^{-9}} = 5.088 \times 10^{14}\ \text{Hz}$

Step 3: Energy difference between excited states = difference in photon energies. …

Why this approach works

When an atom absorbs a photon, the photon’s energy exactly equals the difference between two atomic energy levels. Here, two absorption lines at 589.0 nm and 589.6 nm mean there are two slightly different upper energy levels (the “doublet” arises from spin‑orbit coupling in the sodium atom). The lower level is the same for both transitions. So:

• The frequency of each transition comes directly from $\nu = c / \lambda$.
• The energy difference between the two excited states is simply the difference in the photon energies of the two transitions — because both photons start from the same ground state.

Step‑by‑step calculation

1. Convert wavelengths to metres

The given wavelengths are in nanometres. For calculations in SI units:

2. Find the frequency of each transition

Use the wave equation $c = \nu \lambda$, so $\nu = c / \lambda$. Speed of light $c = 3.00 \times 10^8\ \text{m/s}$.

For the first line:

For the second line:

3. Calculate the photon energy for each transition

Planck’s relation: $E = h\nu$, with $h = 6.626 \times 10^{-34}\ \text{J·s}$.

4. Find the energy difference between the two excited states

Both transitions start from the same lower state. So the difference in photon energies equals the difference in the upper state energies:

That is:

But let’s be more precise using the original numbers without rounding:

First compute the bracket:

$\frac{1}{589.0} - \frac{1}{589.6} = \frac{589.6 - 589.0}{589.0 \times 589.6} = \frac{0.6}{589.0 \times 589.6}$ …