 Solution

For exponential decay with mean lifetime $\tau$, the state amplitude evolves as $\exp [-i(E_0-i\Gamma /2)t/\hslash ]$ and its survival probability is $e^{-\Gamma t/\hslash }$. Hence the energy decay width isThe exceptionally small energy width expresses the long lifetime of this weak decay. Use $\hslash$, not $h$, because $\Gamma$ is an energy width in the complex-energy convention; changing to a frequency convention introduces the appropriate factor of $2\pi$. The quoted lifetime is treated as the mean lifetime. If a half-life were supplied instead, one would first use $\tau =t_{1/2}/\ln 2$. No extra logarithm is needed for the given mean-lifetime interpretation. This calculation uses the source's approximate 15-minute value rather than silently replacing it with a more precise contemporary neutron-lifetime measurement.