Question

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

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

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.

Given Data

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

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\[{\rm{1}}{\rm{.07 \times 1}}{{\rm{0}}^{{\rm{14}}}}\].

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\[{\rm{1}}{\rm{.96 \times 1}}{{\rm{0}}^{{\rm{17}}}}\].