UPSC Combined Geo-Scientist, Mains · 2022 · Mains Paper II
Q1(k). Energy uncertainty of an excited state
Question
From the “uncertainty relation” how would you justify that the uncertainty in measuring the energy of a photoexcited state is greater than that of the ground state?
Answer in brief
ΔE·τ ≥ ℏ/2: the ground state lives for ever (τ → ∞, ΔE → 0); an excited state decays in about 10⁻⁸ s, so its energy is uncertain by ΔE ≥ ℏ/2τ.
Key points
- Energy–time uncertainty: ΔE·Δt ≥ ℏ/2; for a state, Δt is its lifetime τ.
- Ground state: stable, does not decay, τ → ∞ ⇒ ΔE → 0; its energy is sharp.
- Photoexcited state: decays by emission or radiationless steps, τ ≈ 10⁻⁹ to 10⁻⁸ s (singlet); for τ = 10 ns, ΔE ≥ 5.27 × 10⁻²⁷ J.
- Seen as lifetime (natural) broadening: Δν = 1/2πτ = 15.9 MHz for 10 ns, 159 MHz for 1 ns, only 159 Hz for a phosphorescent triplet (τ = 1 ms).
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.

Coversuncertainty principleenergy–time uncertaintyexcited state lifetimenatural line widthlifetime broadening
All parts of Question 1
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- 1(g) Half-life of a second-order reaction5 marks
- 1(h) Brownian movement and Tyndall effect: why only colloids5 marks
- 1(i) Nernst equation for the redox electrode Pt | Cr⁶⁺, Cr³⁺5 marks
- 1(j) Same electrode positive or negative5 marks
- 1(k) Energy uncertainty of an excited state5 marks
- 1(l) Eigenvalues of a Hermitian operator are real5 marks
- 1(m) Rotational constants of ¹⁴N¹⁶O and ¹⁵N¹⁶O5 marks
- 1(n) Photon–molecule interaction in rotating HCl5 marks
- 1(o) Why radiative transitions are drawn vertical5 marks
- 1(p) Energy of one einstein at 600 nm and 300 nm5 marks