What isꝩfor a proton having a mass energy of 938.3MeV accelerated through an effective potential of 1.0TV (teravolt) at Fermilab outside Chicago?

The mass energy of proton is 938.3MeV

Its effective potential is 1.0TV.

In essence, relativistic kinetic energy describes a particle's kinetic energy as the difference between that energy and its rest mass energy. This equation resembles the non-relativistic kinetic energy statement for low velocities.

With the help of the equation (\[{\rm{28}}{\rm{.52}}\]), the relativistic kinetic energy is obtained by:

\[K{E_{rel}}=(\gamma-1)m{c^2}\] …(i)

We have the value of \[\gamma \] as the relativistic constant.

The value of \[m{c^2}\] is said to be the rest mass energy of the particle.

The rest mass of the proton is then obtained as:

\[m{c^2} = 938.3\,{\rm{MeV}}\].

The effective potential to accelerate the proton is then obtained as:

\[V = 1.0\,{\rm{TV}}\].

So, the kinetic energy of the proton is then evaluated as:

\begin{aligned}K{E_{rel{\rm{ }}}}={\rm{eV}}\\ &=1.0\,{\rm{TeV}}\\ &=1.0 \times {10^6}\,{\rm{MeV}}\end{aligned}

Now, with the help of the first equation, we get:

\begin{aligned}\gamma =\frac{{K{E_{rel}}}}{{m{c^2}}}+1\\ &=\frac{{1.0\times {{10}^6}{\rm{MeV}}}}{{938.3\,{\rm{MeV}}}}+1\\=1066.75\\ &= 1067\end{aligned}

Therefore, the proton is: \[\gamma = 1067\].