Q.18
A molecular electronic excited state has a lifetime of 10−9 s.
The uncertainty in measuring the frequency (Hz) of the electronic transition is approximately
Options:
(A) h⁄4π × 109
(B) h⁄4π × 10−9
(C) 1⁄4π × 10−9
(D) 1⁄4π × 109
Uncertainty in Frequency of Electronic Transition
In quantum mechanics and spectroscopy, a finite lifetime of an excited
electronic state causes uncertainty in the energy and frequency of
the transition. This uncertainty is explained using
Heisenberg’s uncertainty principle.
Heisenberg’s Uncertainty Principle
The energy–time uncertainty relation is:
ΔE · Δt ≈ h / 4π
Energy and frequency are related as:
ΔE = hΔν
Derivation
Substituting ΔE = hΔν in the uncertainty relation:
hΔν · Δt ≈ h / 4π
Cancelling h:
Δν ≈ 1 / (4πΔt)
Calculation
Given:
Δt = 10−9 s
Δν ≈ 1 / (4π × 10−9)
Δν ≈ (1 / 4π) × 109 Hz
Correct Answer
Option (D)
Δν ≈ (1 / 4π) × 109 Hz
Summary
| Quantity | Value / Relation |
|---|---|
| Lifetime (Δt) | 10−9 s |
| Uncertainty Relation | ΔEΔt = h/4π |
| Frequency Uncertainty | (1 / 4π) × 109 Hz |
Conclusion
A shorter lifetime of an excited electronic state results in a larger
uncertainty in frequency. Using Heisenberg’s uncertainty principle,
the correct uncertainty is obtained as:
(1 / 4π) × 109 Hz
Hence, the correct answer is Option (D).
