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The Heisenberg uncertainty principle is given by,

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

Here\(\Delta E\)is the uncertainty in energy,\(\Delta t\)is uncertainty in time and\(h\)is the Planck鈥檚 constant.

(a)

Heisenberg uncertainty principle for energy states that:

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

Substitute\({\rm{1}}{{\rm{0}}^{{\rm{ - 43}}}}{\rm{s}}\)for\(\Delta t\)in the above equation,

So, we obtain:

\(\begin{align}\Delta E \approx \frac{h}{{4\pi \Delta t}}\\ &= \frac{{{\rm{6}}{\rm{.626 \times 1}}{{\rm{0}}^{{\rm{ - 43}}}}{\rm{ Js}}}}{{{\rm{4\pi \times 1}}{{\rm{0}}^{{\rm{ - 43}}}}{\rm{ s}}}}\\ &= {\rm{5}}{\rm{.27 \times 1}}{{\rm{0}}^{\rm{8}}}{\rm{J}}\end{align}\)

Therefore, the uncertainty is\({\rm{5}}{\rm{.27 \times 1}}{{\rm{0}}^{\rm{8}}}{\rm{ J}}\).

Convert it into \({\rm{eV}}\)

\(\begin{align}\Delta E &= \left( {5.2728 \times {{10}^8}\,{\rm{J}}} \right)\left( {\frac{{1\,{\rm{eV}}}}{{1.6 \times {{10}^{ - 19}}\,{\rm{J}}}}} \right)\\ &= 3 \times {10^{27}}\,{\rm{eV}}\\ &= 3 \times {10^{18}}\,{\rm{GeV}}\end{align}\)

Therefore the uncertainty in energy is \(3 \times {10^{18}}\,{\rm{GeV}}\).

(b)

The ratio is

\(\begin{align}\frac{{\Delta E}}{E} &= \frac{{3 \times {{10}^{18}}\,{\rm{GeV}}}}{{{{10}^{18}}\,{\rm{GeV}}}}\\ &= 0.33\end{align}\)

Due to its high energy, such a particle might be involved in the unification of the strong and electroweak forces, but it would be unable to detect with present techniques due to its brief life in such an energy state.

Therefore, due to the Heisenberg uncertainty principle predicts such a particle's existence will be too brief, this theoretical energy threshold precludes the detection of such a particle.