Two protons are situated at a distance of 100 fermi from each other. The potential energy of this system will be in eV

The correct option is 1.44 x 10⁴.

To find the potential energy of the system of two protons, we can use Coulomb's formula for electrostatic potential energy ($U$):

$U=\frac{k\cdot q_1\cdot q_2}{r}$

Where:
• $k$ is Coulomb's constant, approximately equal to $9\times {10}^9{\text{N·m}}^2/{\text{C}}^2$.
• $q_1$ and $q_2$ are the charges of the two protons, each having a charge equal to the elementary charge, $e=1.6\times {10}^{-19}\text{C}$.
• $r$ is the distance between the protons. Here, $r=100\text{fermi}$. Since $1\text{fermi}={10}^{-15}\text{m}$, we convert the distance to meters:
$r=100\times {10}^{-15}\text{m}={10}^{-13}\text{m}$.

The potential energy calculated using Coulomb's law will be in Joules (J). To find the energy in electron-volts (eV), we divide the energy in Joules by the elementary charge $e$ ($1.6\times {10}^{-19}\text{J/eV}$):

$U_{\text{eV}}=\frac{U}{e}=\frac{k\cdot e^2}{r\cdot e}=\frac{k\cdot e}{r}$

Substituting the known values into the equation:

$U_{\text{eV}}=\frac{(9\times {10}^9)\times (1.6\times {10}^{-19})}{{10}^{-13}}$

$U_{\text{eV}}=\frac{14.4\times {10}^{-10}}{{10}^{-13}}$

$U_{\text{eV}}=14.4\times {10}^3\text{eV}=1.44\times {10}^4\text{eV}$

Thus, the potential energy of the system is $1.44\times {10}^4\text{eV}$.