The mass of a theoretical particle that may be associated with the unification of the electroweak and strong forces is\[{\rm{1}}{{\rm{0}}^{{\rm{14}}}}{\rm{ GeV/}}{{\rm{c}}^{\rm{2}}}\]. (a) How many proton masses is this? (b) How many electron masses is this? (This indicates how extremely relativistic the accelerator would have to be in order to make the particle, and how large the relativistic quantity γ would have to be.)

Short Answer

Expert verified

a) 1014GeV/c2is equal to 1.07 x 1014 proton masses.

b)1014GeV/c2 is equal to 1.96 x 1017 electron masses.

Step by step solution

01

Concept Introduction

Charged particles, such as protons or electrons, are propelled at high speeds near the speed of light by an accelerator. They are subsequently crushed against other particles flowing in the opposite direction or against a target. Physicists can examine the realm of the endlessly small by investigating these collisions.

02

Given Data

Mass of the theoretical particle is m = 1014GeV/C2.

03

Calculating the proton masses

a)

Dividing the theoretical particle's rest mass\(\left( m \right)\)by the rest masses of the proton\(\left( {{m_p}} \right)\), we get-

\[\frac{m}{{{m_p}}} = \frac{{{{10}^{14}}\;{\rm{GeV/}}{{\rm{c}}^{\rm{2}}}}}{{938 \times {{10}^{ - 3}}\;{\rm{GeV/}}{{\rm{c}}^{\rm{2}}}}} = 1.07 \times {10^{14}}\]

Therefore, the number of proton masses is\[{\rm{1}}{\rm{.07 \times 1}}{{\rm{0}}^{{\rm{14}}}}\].

04

Calculating the electron masses

b)

Dividing the theoretical particle's rest mass\(\left( m \right)\)by the rest masses of the proton\(\left( {{m_p}} \right)\), we get-

\[\frac{m}{{{m_e}}} = \frac{{{{10}^{14}}\;{\rm{GeV/}}{{\rm{c}}^{\rm{2}}}}}{{0.511 \times {{10}^{ - 3}}\;{\rm{GeV/}}{{\rm{c}}^{\rm{2}}}}} = 1.96 \times {10^{17}}\]

Therefore, the number of electron masses is\[{\rm{1}}{\rm{.96 \times 1}}{{\rm{0}}^{{\rm{17}}}}\].

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