Question

The average particle energy needed to observe unification of forces is estimated to be\({\rm{1}}{{\rm{0}}^{{\rm{19}}}}{\rm{ GeV}}\). (a) What is the rest mass in kilograms of a particle that has a rest mass of\({\rm{1}}{{\rm{0}}^{{\rm{19}}}}{\rm{ GeV/ }}{{\rm{c}}^{\rm{2}}}\)? (b) How many times the mass of a hydrogen atom is this?

Short answer

(a) The rest mass in kilogram of a particle is \(1.7 \times {10^{ - 8}}\,{\rm{kg}}\).

(b) The rest mass of particles as a multiple of mass of the hydrogen is \({10^{19}}\).

Step-by-step solution

Energy expression.

In case of the annihilation if the mass gets converted into the energy. If we get value of total mass destroyed then the expression for the energy is obtained by,

\(E = m{c^2}\)

Here\(E\)is the energy,\(m\)is the mass and\(c\)is the speed of the light.

Evaluating the mass

It is given that the energy needed for the unification of force is \({10^{19}}\,{\rm{GeV}}\).

The formulae for the rest mass from the energy expression is,

\(m = \frac{E}{{{c^2}}}\)

It is given that the rest mass is \({10^{19}}\,{\rm{GeV/}}{{\rm{c}}^{\rm{2}}}\), convert it into kg,

\(\begin{array}{c}m = \frac{{\left( {{{10}^{19}}\,{\rm{GeV}}} \right)\left( {1.6 \times {{10}^{ - 19}}\,{\rm{C}}} \right)\left( {{{10}^9}\,{\rm{KeV/GeV}}} \right)}}{{{{\left( {3.00 \times {{10}^8}\,{\rm{m/s}}} \right)}^2}}}\\ = 1.7 \times {10^{ - 8}}\,{\rm{kg}}\end{array}\)

Therefore the rest mass in kilogram of a particle is \(1.7 \times {10^{ - 8}}\,{\rm{kg}}\).

Evaluating the number of times of hydrogen atom

The ratio of rest mass to the mass of the hydrogen is,

\(\begin{array}{c}{\rm{Ratio}} = \frac{{{\rm{1}}{\rm{.78 \times 1}}{{\rm{0}}^{{\rm{ - 8}}}}{\rm{ kg}}}}{{{\rm{1}}{\rm{.67 \times 1}}{{\rm{0}}^{{\rm{ - 27}}}}{\rm{ kg}}}}\\{\rm{ }} = {\rm{ 1}}{{\rm{0}}^{{\rm{19}}}}\end{array}\)

Therefore, the rest mass of particles as a multiple of mass of the hydrogen is \({10^{19}}\).