Question

A virtual particle having an approximate mass of \[{\rm{1}}{{\rm{0}}^{{\rm{14}}}}{\rm{GeV/}}{{\rm{c}}^{\rm{2}}}\]may be associated with the unification of the strong and electroweak forces. For what length of time could this virtual particle exist (in temporary violation of the conservation of mass-energy as allowed by the Heisenberg uncertainty principle)?

Short answer

This lifespan of the particle is \[3.3 \times {10^{ - 39}}\;{\rm{s}}\].

Step-by-step solution

Definition of Heisenberg’s uncertainty principle

The Uncertainty Principle of Heisenberg states that it is impossible to precisely compute energy transferred and the time taken simultaneously. There will at least be an uncertainty equal toh/4π . Here, h is Plank’s constant.

Given data

Mass of the particle is m = 10¹⁴MeV/c².

Finding at what length of time the particle could exist

Using the uncertainty equation as a guide,

\[{\rm{\Delta E\Delta t = }}\frac{{\rm{h}}}{{{\rm{4\pi }}}}\],

we can get the duration of existence\[\left( {{\rm{\Delta t}}} \right)\]and uncertainty in energy\[{\rm{\Delta E = \Delta m}}{{\rm{c}}^{\rm{2}}}\]

\[\begin{array}{c}\Delta t = \frac{h}{{4\pi \Delta m{c^2}}}\\ = \frac{{\left( {6.63 \times {{10}^{ - 34}}\;{\rm{J}}{\rm{.s}}} \right)}}{{4\pi \times \left( {{{10}^{14}} \times {{10}^9} \times 1.6 \times {{10}^{ - 19}}\;{\rm{kg}}} \right)}}\\ = 3.3 \times {10^{ - 39}}\;{\rm{s}}\end{array}\]

Therefore,this particle exists for \[3.3 \times {10^{ - 39}}\;{\rm{s}}\]