Question

The Large Hadron Collider accelerates protons to a total energy of $6.5\mathrm{TeV}$per proton. ( $1\mathrm{TeV}=1$teraelectron volt $={10}^{12}\mathrm{eV}$.) What is the speed of a proton that has this energy? Give your answer as fraction of $c$, using as many significant figures as needed.

Short answer

The required speed (as a fraction of $c$) is $v=0.9999999895·c$

Step-by-step solution

Given Information

Let us start the solution with the expression for relativistic total energy
$E=\frac{m{c}^2}{\sqrt{1-\frac{v^2}{c^2}}}$

Calculation

$\begin{array}{rcl}{E}^2\left(1-\frac{v^2}{c^2}\right)& =& m^2{c}^4\\ E^2\frac{v^2}{c^2}& =& E^2-m^2{c}^4\\ v& =& \sqrt{c^2-\frac{m^2{c}^6}{E^2}}\\ v& =& c\sqrt{1-\frac{m^2{c}^4}{E^2}}\\ v& =& c·\sqrt{1-\frac{{\left(1.67·{10}^{-27}\right)}^2·{\left(3·{10}^8\right)}^4}{6.5·{10}^{12}·1.6·{10}^{-19}}}\\ v& =& 0.9999999895·c\end{array}$