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The Large Hadron Collider accelerates protons to a total energy of 6.5TeVper proton. ( 1TeV=1teraelectron volt =1012eV.) 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

Expert verified

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

Step by step solution

01

Given Information

Let us start the solution with the expression for relativistic total energy
E=mc21-v2c2

02

Calculation

E21-v2c2=m2c4E2v2c2=E2-m2c4v=c2-m2c6E2v=c1-m2c4E2v=c·1-1.67·10-272·3·10846.5·1012·1.6·10-19v=0.9999999895·c

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