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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

Expert verified

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

Step by step solution

01

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.

02

Given data

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

03

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 \[3.3 \times {10^{ - 39}}\;{\rm{s}}\]

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Most popular questions from this chapter

What length track does a \[{{\rm{\pi }}^{\rm{ + }}}\]traveling at 0.100c leave in a bubble chamber if it is created there and lives for \[{\rm{2}}{\rm{.60 \times 1}}{{\rm{0}}^{{\rm{ - 8}}}}{\rm{\;s}}\]? (Those moving faster or living longer may escape the detector before decaying.)

(a) Show that the conjectured decay of the proton, \({\rm{p}} \to {\pi ^{\rm{0}}}{\rm{ + }}{{\rm{e}}^{\rm{ + }}}\), violates conservation of baryon number and conservation of lepton number.

(b) What is the analogous decay process for the antiproton?

The principal decay mode of the sigma zero is \[{{\rm{\Sigma }}^{\rm{0}}}{\rm{ }} \to {\rm{ }}{{\rm{\Lambda }}^{\rm{0}}}{\rm{ + \gamma }}\]. (a) What energy is released? (b) Considering the quark structure of the two baryons, does it appear that the \[{{\rm{\Sigma }}^{\rm{0}}}\]is an excited state of the \[{{\rm{\Lambda }}^{\rm{0}}}\]? (c) Verify that strangeness, charge, and baryon number are conserved in the decay. (d) Considering the preceding and the short lifetime, can the weak force be responsible? State why or why not.

The sigma-zero particle decays mostly via the reaction \[{{\rm{\Sigma }}^{\rm{0}}} \to {{\rm{\Lambda }}^{\rm{0}}}{\rm{ + \gamma }}\]. Explain how this decay and the respective quark compositions imply that the \[{{\rm{\Sigma }}^{\rm{0}}}\]is an excited state of the\[{{\rm{\Lambda }}^{\rm{0}}}\].

The primary decay mode for the negative pion \({\pi ^{\rm{ - }}} \to {{\rm{\mu }}^{\rm{ - }}}{\rm{ + }}{{\rm{\bar \upsilon }}_{\rm{\mu }}}\).

(a) What is the energy release in \({\rm{MeV}}\) in this decay?

(b) Using conservation of momentum, how much energy does each of the decay products receive, given the \({\pi ^{\rm{ - }}}\) is at rest when it decays? You may assume the muon antineutrino is massless and has momentum \(p = \frac{{{E_\nu }}}{c}\), just like a photon.

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