Example 29.9 from College Physics, 29.7 Probability: The Heisenberg Uncertainty Principle

Heisenberg Uncertainty Principle for Energy and Time for an Atom

An atom in an excited state temporarily stores energy. If the lifetime of this excited state is measured to be ${1.0×10}^{- 10} s$ , what is the minimum uncertainty in the energy of the state in eV?

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
The minimum uncertainty in energy $Δ E$ is found by using the equals sign in $Δ E Δ t \geq h /4 π$ and corresponds to a reasonable choice for the uncertainty in time. The largest the uncertainty in time can be is the full lifetime of the excited state, or $Δ t = {1.0×10}^{- 10} s$ .
Solution
Solving the uncertainty principle for $Δ E$ and substituting known values gives
$Δ E = \frac{h}{4πΔt} = \frac{6 . 63 \times 10^{–34} J \cdot s}{4π ({1.0×10}^{–10} s)} = 5 . 3 \times 10^{–25} J.$
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Now converting to eV yields

$Δ E = (5.3 \times 10^{–25} J) (\frac{1 eV}{1 . 6 \times 10^{–19} J}) = 3 . 3 \times 10^{–6} eV .$
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Discussion
The lifetime of $10^{- 10} s$ is typical of excited states in atoms—on human time scales, they quickly emit their stored energy. An uncertainty in energy of only a few millionths of an eV results. This uncertainty is small compared with typical excitation energies in atoms, which are on the order of 1 eV. So here the uncertainty principle limits the accuracy with which we can measure the lifetime and energy of such states, but not very significantly.

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