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Question

An atom typically exists in an excited state for about Δt=108 s. Estimate the uncertainity in frequency Δν in the frequency of emitted photons when an atom makes a transition from an excited state with the simultaneous emission of a photon with an average frequency of νav = 7.1×1014 Hz. Calculate the ratio of uncertainity in frequency to the average frequency.

A
1.12×1010
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B
2.11×1010
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C
2.11×108
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D
1.12×108
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Solution

The correct option is A 1.12×1010
We know,
From Heisenberg uncertainty principle,
ΔE.(Δt)h4π
Again, from Plack's quantum theory, we can say E=hΔν
Combining the relations, we get, h(Δ.ν)Δth4π
Or (Δν)h4π.Δt.h
(Δν) = 14π×1088×106 Hz
So, Δννav = 8×1067.1×1014 = 1.12×108

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