Question

The decay energy of a short-lived nuclear excited state has an uncertainty of 2.0 eV due to its short lifetime. What is the smallest lifetime it can have?

Short answer

The minimum lifetime of the particle will be\(1.6 \times {10^{ - 16}}\;{\rm{s}}\).

Step-by-step solution

Determine the formulas

The equation (\({\rm{29}}{\rm{.49}}\)), Heisenberg's uncertainty principle for simultaneous measurements of energy and time in equation form is:

\({\bf{\Delta E\Delta t}} \ge \frac{{\bf{h}}}{{{\bf{4\pi }}}}\)

The value we have is:\(h = 6.626 \times {10^{ - 34}}{\rm{ J}} \cdot {\rm{s}}\)which is Planck's constant.

The value of\({\rm{\Delta E}}\)is said to be the uncertainty in energy.

The value of\({\rm{\Delta t}}\)is said to be the uncertainty in time.

With the help of the first equation, the minimum uncertainty in energy is:

\({\bf{\Delta E}} \ge \frac{{\bf{h}}}{{{\bf{4\pi \Delta t}}}}\)

Find the smallest lifetime of nuclear excited state

Determine the uncertainty in decay energy:

\(\begin{aligned}\Delta E &= 2.0\;{\rm{eV}}\\ &= \left( {2.0\;{\rm{eV}}} \right)\left( {1.6 \times {{10}^{ - 19}}\;\frac{{\rm{J}}}{{{\rm{eV}}}}} \right)\\ &= 3.2\;\;{\rm{J}}\end{aligned}\)

As a result, life expectancy is unknown.

\(\begin{aligned}\Delta t &= \frac{h}{{4\pi \Delta E}}\\ &= \frac{{6.67 \times {{10}^{ - 34}}\;{\rm{J}} \cdot {\rm{s}}}}{{4\pi \left( {3.2 \times {{10}^{ - 19}}\;{\rm{J}}} \right)}}\\ &= 1.6 \times {10^{ - 16}}\;{\rm{s}}\end{aligned}\)

Hence, the minimum lifetime of the particle will be \(1.6 \times {10^{ - 16}}\;{\rm{s}}\).