Question

Protons in an accelerator at the Fermi National Laboratory near chicago are accelerated to a total energy that is 400 times their rest energy.
(a) What is the speed of these protons in terms of c?
(b) What is their kinetic energy in MeV?

Answer 1

(a) 0.99999687c

The relationship between total energy (E) and rest energy ($E_0$) is

$E=\gamma E_0$

where $\gamma$ is the relativistic factor. In this problem, we know that the total energy is 400 times the rest energy: this means that

$\gamma=400$

The formula for the relativistic factor is

$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

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:

$1-(\frac{v}{c})^2=\frac{1}{\gamma^2}\\\frac{v}{c}=\sqrt{1-\frac{1}{\gamma^2}}=\sqrt{1-\frac{1}{400^2}}=0.99999687$

(b) $3.75\cdot 10^5 MeV$

Given the rest mass of a proton:

$m_0=1.67\cdot 10^{-27}kg$

its rest energy is the energy equivalent to this mass:

$E_0=m_0 c^2 = (1.67\cdot 10^{-27}kg)(3\cdot 10^8 m/s)^2=1.5\cdot 10^{-10} J$

The protons in the Fermi Laboratory have energy that is 400 times their rest energy, so their total energy is

$E=400 E_0$

The total energy is also sum of rest energy and kinetic energy:

$E=E_0+K$

So the kinetic energy is

$K=E-E_0=400E_0 - E_0 = 399 E_0 = 399(1.5\cdot 10^{-10}J)=6.0\cdot 10^{-8} J$

And since the conversion factor is

$1 eV = 1.6\cdot 10^{-19} J$

This kinetic energy converted into MeV ($10^6 eV$) is

$K=\frac{6.0\cdot 10^{-8}J}{1.6\cdot 10^{-19}J/eV \cdot 10^6 eV/MeV}=3.75\cdot 10^5 MeV$