Skip to content
Exercises · 2.50

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.

Andhra Pradesh BieapTextbookSubjective· 2mImportance★★★★★est
49% · 68/140 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The problem uses the photon energy relation E=hνE = h\nu to find the frequencies of two closely spaced spectral lines (the sodium D‑doublet) and then the tiny energy gap between the two upper excited states that produce them. The frequencies are 5.09×10145.09 \times 10^{14} Hz and 5.086×10145.086 \times 10^{14} Hz; the energy difference between the excited states is 3.31×10−223.31 \times 10^{-22} J.

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 ν=c/λ\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:

λ1=589.0 nm=589.0×10−9 m\lambda_1 = 589.0\ \text{nm} = 589.0 \times 10^{-9}\ \text{m}

λ2=589.6 nm=589.6×10−9 m\lambda_2 = 589.6\ \text{nm} = 589.6 \times 10^{-9}\ \text{m}

2. Find the frequency of each transition

Use the wave equation c=νλc = \nu \lambda, so ν=c/λ\nu = c / \lambda. Speed of light c=3.00×108 m/sc = 3.00 \times 10^8\ \text{m/s}.

For the first line:

ν1=3.00×108589.0×10−9=5.093×1014 Hz\nu_1 = \frac{3.00 \times 10^8}{589.0 \times 10^{-9}} = 5.093 \times 10^{14}\ \text{Hz}

For the second line:

ν2=3.00×108589.6×10−9=5.088×1014 Hz\nu_2 = \frac{3.00 \times 10^8}{589.6 \times 10^{-9}} = 5.088 \times 10^{14}\ \text{Hz}

Tip

Notice that the longer wavelength (589.6 nm) gives the lower frequency — wavelength and frequency are inversely proportional. This is a quick sanity check.

3. Calculate the photon energy for each transition

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

E1=(6.626×10−34)(5.093×1014)=3.375×10−19 JE_1 = (6.626 \times 10^{-34})(5.093 \times 10^{14}) = 3.375 \times 10^{-19}\ \text{J}

E2=(6.626×10−34)(5.088×1014)=3.371×10−19 JE_2 = (6.626 \times 10^{-34})(5.088 \times 10^{14}) = 3.371 \times 10^{-19}\ \text{J}

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:

ΔE=E1−E2=(3.375−3.371)×10−19 J=0.004×10−19 J\Delta E = E_1 - E_2 = (3.375 - 3.371) \times 10^{-19}\ \text{J} = 0.004 \times 10^{-19}\ \text{J}

That is:

ΔE=4.0×10−22 J\Delta E = 4.0 \times 10^{-22}\ \text{J}

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

ΔE=hc(1λ1−1λ2)\Delta E = hc \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right)

ΔE=(6.626×10−34)(3.00×108)(1589.0×10−9−1589.6×10−9)\Delta E = (6.626 \times 10^{-34})(3.00 \times 10^8) \left( \frac{1}{589.0 \times 10^{-9}} - \frac{1}{589.6 \times 10^{-9}} \right)

First compute the bracket:

1589.0−1589.6=589.6−589.0589.0×589.6=0.6589.0×589.6\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} …

Start your 14-day free trial to unlock the full solution →

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests
Start 14-day free trial →See Plans & Pricing →

Full access for 14 days. No credit card required.

←Previous question · exercises2.49 — Lifetimes of the molecules in the excited states are often measured by…→Next question · exercises2.51 — The work function for caesium atom is 1.9 eV. Calculate (a) the thresh…