Question

The decay energy of a short-lived particle has an uncertainty of\({\bf{1}}{\bf{.0 MeV}}\)due to its short lifetime. What is the smallest lifetime it can have?

Short answer

The smallest lifetime is obtained as \(3.3 \times {10^{ - 22}}{\rm{ s}}\).

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:

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

Evaluate the smallest lifetime

Determine the energy value in joule.

\(\begin{aligned}\Delta E &= 1.0\;{\rm{MeV}}\\ &= 1.6 \times {10^{ - 13}}{\rm{ J}}\end{aligned}\)

Substitute the values, the uncertainty in time is obtained as:

\(\begin{aligned}\Delta t &= \frac{h}{{4\pi \Delta E}}\\ &= \frac{{6.67 \times {{10}^{ - 34}}\;{\rm{J}} \cdot {\rm{s}}}}{{4\pi \left( {1.6 \times {{10}^{ - 13}}\;{\rm{J}}} \right)}}\\ &= 3.3 \times {10^{ - 22}}{\rm{ s}}\end{aligned}\)

Therefore, the lifetime of the particle is \(3.3 \times {10^{ - 22}}{\rm{ s}}\).