In nuclear fission, a nucleus splits roughly in half. (a) What is the potential \(2.00 \times {10^{ - 14}}{\rm{ }}m\) from a fragment that has \(46\) protons in it? (b) What is the potential energy in \(MeV\) of a similarly charged fragment at this distance?

The potential at a distance\(r{\rm{ }} = {\rm{ }}2.00{\rm{ }}x{\rm{ }}{10^{ - 14}}{\rm{ }}m\)from a fragment of the nucleus has\({\rm{46}}\)protons.

The charge of a single proton is:

\(\begin{array}{c}{q_p}{\rm{ }} = {\rm{ }} + e{\rm{ }}\\ = {\rm{ }} + 1.60{\rm{ }} \times {\rm{ }}{10^{ - 19}}{\rm{ }}C\end{array}\)

In the part (a), we are suppose to determine the potential a distance\(r\)from the fragment.

In part (b), we determine the potential energy of a similarly charged fragment at a distance \(r\) from the original fragment.

A physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them, is referred to as an electric field.

The electric potential due to a Single charge is : To determine the electric potential due to a single source charge at a specific location, we place a test charge\({\rm{q}}\)at that location and then determine the electric potential energy\({U_{Qq}}\)of a system containing the test charge and the source charge that creates the field . The electric potential at that location equals to the ratio of:

\(\begin{array}{c}V = \frac{{{U_{Qq}}}}{q}\\ = \frac{{kQ}}{r}\end{array}\)…………….(I)

Here, the value is the Coulomb's constant, a proportionality constant. The joule/coulomb is referred to as the unit of electric potential and is referred to as the volt\((V)\).

Electric Potential Energy: If a charged particle\(q\)is positioned at a location where the electric potential of a charged object\(V\)is claimed to be, the particle-object system's electric potential energy is:

\(Ue{\rm{ }} = {\rm{ }}qV\)…………...(II)

Electron Volt: It is the energy given to a fundamental charge accelerated through a potential difference of\(\left( {1{\rm{ }}V} \right)\). It is in equation form as:

\(\begin{array}{c}1{\rm{ }}eV{\rm{ }} = {\rm{ }}\left( {1.60{\rm{ }}x{\rm{ }}10 - 19{\rm{ }}C} \right)\left( {1{\rm{ }}V} \right)\\ = {\rm{ }}1.60{\rm{ }}x{\rm{ }}10 - 19{\rm{ }}J\end{array}\)…………………….(III)

\(V = \frac{{kQ}}{r}\)

The value of\(Q = n{p_q}\), is the charge of\(46\)protons as:

\(V = \frac{{k(46{q_p})}}{r}\)

Entering the values, we then obtain:

\(\begin{array}{c}V = \frac{{46(8.99 \times {{10}^9}N.{m^2}/C)(1.60 \times {{10}^{ - 19}}C)}}{{2.00 \times {{10}^{ - 14}}m}}\\ = 3.31 \times {10^6}V\end{array}\)

Therefore, the electric potential from a distance is: \(V = 3.31 \times {10^6}{\rm{ }}V\).

\(Ue{\rm{ }} = {\rm{ }}qV\)

The value of\(\begin{array}{c}q{\rm{ }} = {\rm{ }}46{q_p}\\ = {\rm{ }}46e\end{array}\) which is the charge of the fragment as:

\(Ue{\rm{ }} = {\rm{ }}46eV\)

Now, entering the values for\(e\)and\(V\), and then converting \(Ue\)into \(MeV\)by using the conversion factor in the third equation as:

\(\begin{array}{c}{U_e} = 46(1.60 \times {10^{ - 19}}C)(3.31 \times {10^6}V)(\frac{{1eV}}{{1.60 \times {{10}^{ - 19}}C.V}})(\frac{{1MeV}}{{{{10}^6}eV}})\\ = 46(3.31eV)\\ = 152MeV\end{array}\)

Therefore, the potential energy of the system containing the original and similarly charged fragments as: \({U_e} = 152MeV\).