(a) 0.99999687c
The relationship between total energy (E) and rest energy ([tex]E_0[/tex]) is
[tex]E=\gamma E_0[/tex]
where [tex]\gamma[/tex] is the relativistic factor. In this problem, we know that the total energy is 400 times the rest energy: this means that
[tex]\gamma=400[/tex]
The formula for the relativistic factor is
[tex]\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
where v is the speed of the proton and c is the speed of light. By isolating the term v/c, we find the speed of the protons in terms of c:
[tex]1-(\frac{v}{c})^2=\frac{1}{\gamma^2}\\\frac{v}{c}=\sqrt{1-\frac{1}{\gamma^2}}=\sqrt{1-\frac{1}{400^2}}=0.99999687[/tex]
(b) [tex]3.75\cdot 10^5 MeV[/tex]
Given the rest mass of a proton:
[tex]m_0=1.67\cdot 10^{-27}kg[/tex]
its rest energy is the energy equivalent to this mass:
[tex]E_0=m_0 c^2 = (1.67\cdot 10^{-27}kg)(3\cdot 10^8 m/s)^2=1.5\cdot 10^{-10} J[/tex]
The protons in the Fermi Laboratory have energy that is 400 times their rest energy, so their total energy is
[tex]E=400 E_0[/tex]
The total energy is also sum of rest energy and kinetic energy:
[tex]E=E_0+K[/tex]
So the kinetic energy is
[tex]K=E-E_0=400E_0 - E_0 = 399 E_0 = 399(1.5\cdot 10^{-10}J)=6.0\cdot 10^{-8} J[/tex]
And since the conversion factor is
[tex]1 eV = 1.6\cdot 10^{-19} J[/tex]
This kinetic energy converted into MeV ([tex]10^6 eV[/tex]) is
[tex]K=\frac{6.0\cdot 10^{-8}J}{1.6\cdot 10^{-19}J/eV \cdot 10^6 eV/MeV}=3.75\cdot 10^5 MeV[/tex]
