Question

(a) What is the uncertainty in the energy released in the decay of a \({{\rm{\pi }}^{\rm{0}}}\)due to its short lifetime?

(b) What fraction of the decay energy is this, noting that the decay mode is \({{\bf{\pi }}^{\bf{0}}} \to {\bf{\gamma }}{\rm{ }} + {\rm{ }}{\bf{\gamma }}\) (so that all the \({\rm{\pi ^0}}\)mass is destroyed)?

Short answer

a) The uncertainty in decay energy is\({\rm{3}}{\rm{.9}}\;{\rm{eV}}\).

b)The fractional uncertainty in decay energy is \(2.9 \times {10^{ - 8}}\).

Step-by-step solution

Concept Introduction

According to the law of conservation of energy, in a decay process, the total energy of all the particles before the decay must be equal to the total energy of the particles after the decay.

Calculating the energy

a)

We can get the uncertainty in the energy using the equation

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

For uncertainty in time\(\Delta t = 8.4 \times {10^{ - 17}}\;{\rm{s}}\), we have

\(\begin{array}{c}\Delta E & = \frac{h}{{4\pi \Delta t}}\\ &= \frac{{\left( {4.14 \times {{10}^{ - 24}}} \right)\;{\rm{GeV/}}{{\rm{c}}^{\rm{2}}}}}{{4\pi \times \left( {8.4 \times {{10}^{ - 17}}} \right)\;{\rm{s}}}}\\ &= 3.9\;{\rm{eV}}\end{array}\)

Therefore, the required solution is \({\rm{3}}{\rm{.9}}\;{\rm{eV}}\).

Calculating the decay energy

b)

For fractional uncertainty, we divide the energy uncertainty by the decay energy of the\({{\rm{\pi }}^{\rm{0}}}\)particle.

\(\frac{{\Delta E}}{{{E_\pi }}} = \frac{{3.9\;\;{\rm{eV}}}}{{135 \times {{10}^6}\;{\rm{eV}}}} = 2.9 \times {10^{ - 8}}\)

Therefore, the fractional uncertainty is \(2.9 \times {10^{ - 8}}\).