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Heisenberg's Uncertainty Principle

An atom typic...

Question

An atom typically exists in an excited state for about $Δ t = 10^{- 8}$ s. Estimate the uncertainity in frequency $Δ ν$ 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 $ν_{a v}$ = $7.1 \times 10^{14}$ Hz. Calculate the ratio of uncertainity in frequency to the average frequency.

1.12 \times 10^{- 10}

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2.11 \times 10^{- 10}

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2.11 \times 10^{- 8}

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1.12 \times 10^{- 8}

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Solution

The correct option is A $1.12 \times 10^{- 10}$
We know,
From Heisenberg uncertainty principle,
$Δ E . (Δ t) \geq \frac{h}{4 π}$
Again, from Plack's quantum theory, we can say $E = h Δ ν$
Combining the relations, we get, $h (Δ . ν) Δ t \geq \frac{h}{4 π}$
Or $(Δ ν) \geq \frac{h}{4 π . Δ t . h}$
$\Rightarrow (Δ ν)$ = $\frac{1}{4 π \times 10^{- 8}} \approx 8 \times 10^{6}$ Hz
So, $\frac{Δ ν}{ν_{a v}}$ = $\frac{8 \times 10^{6}}{7.1 \times 10^{14}}$ = $1.12 \times 10^{- 8}$

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Similar questions

Δ t = 10^{- 8}

Δ ν

ν_{a v}

7.1 \times 10^{14}

1.0 \times 10^{- 8} s

1.0 \times 10^{- 8} s

10^{- 8} s

n = 2

n = 2

a_{0} = 5.3 \times 10^{- 11} m)

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