Question

An atom in an excited state temporarily stores energy. If the lifetime of this excited state is measured to be $1.0\times {10}^{-10}s$ then the minimum uncertainty in the energy of the state is $X\times {10}^{-6}$ eV. Then X is . Take $h=6.6\times {10}^{-34}Jsand\pi =3$.

Answer 1

The largest uncertainty in time can be the full lifetime of the excited state, or $△t=1.0\times {10}^{-10}$ s.

The maximum uncertainty in time leads to the minimum uncertainty in energy.

Solving the uncertainty principle for $△E$ and substituting known values gives
$△E=\frac{h}{4\pi △t}=\frac{6.6\times {10}^{-34}J.s}{4\pi (1.0\times {10}^{-10}\text{s})}=5.5\times {10}^{-25}J$

Now converting to eV yields
$△E=5.5\times {10}^{-25}\times \frac{1eV}{1.6\times {10}^{-19}J}=3.43\times {10}^{-6}$ eV