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A sodium atom is in one of the states labeled 'Lowest excited levels'. It remains in that state for an average time of 10^–8 sec, before it makes a transition back to a ground state. What is the uncertainty in energy of that excited state

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6.56 x 10^–8 eV

2 x 10^–8 eV

10^–8 eV

8 x 10^–8 eV

answer is A.

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Detailed Solution

The average time that the atom spends in this excited state is equal to Δt, so by using $\Delta E.\,\Delta t = \frac{h}{{2\pi }}$

$\Rightarrow Uncertainty{\text{ }}in{\text{ }}energy = \frac{{h/2\pi }}{{\Delta t}}$

$= \frac{{6.6 \times {{10}^{ - 34}}}}{{2 \times 3.14 \times {{10}^{ - 8}}}} = 1.05 \times {10^{ - 26}}J = 6.56 \times {10^{ - 8}}eV$

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