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Q13PE

Expert-verifiedFound in: Page 1238

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**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?**

(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 \({10^{19}}\).

**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.**

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 \(1.7 \times {10^{ - 8}}\,{\rm{kg}}\).

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 \({10^{19}}\).

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