The potential energy of a pair of nucleons is minimum when they are about 0.8 fm apart. Two protons are placed at this separation. Their electrostatic potential energy is
(Take \(\frac{1}{4\pi\varepsilon_0}=9\times10^{9}\ \mathrm{N\,m^{2}\,C^{-2}}\), \(e=1.6\times10^{-19}\ \mathrm{C}\) and \(1\ \mathrm{MeV}=1.6\times10^{-13}\ \mathrm{J}\).)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1.8 MeV
Two protons repel each other by the Coulomb force. For two charges \( q_1 \) and \( q_2 \) at a distance \( r \), the electrostatic potential energy is \( U=q_1q_2/4\pi\varepsilon_0r \). For two protons, each charge is \( e \).
Given:
\( r=0.8\ \mathrm{fm}=0.8\times10^{-15}\ \mathrm{m} \)
\( e=1.6\times10^{-19}\ \mathrm{C} \)
\( \begin{aligned} U &= \frac{1}{4\pi\varepsilon_0}\frac{e^{2}}{r} \\ &= \frac{9\times10^{9}\times(1.6\times10^{-19})^{2}}{0.8\times10^{-15}} \\ &= 2.88\times10^{-13}\ \mathrm{J} \end{aligned} \)
In MeV,
\( \begin{aligned} U &= \frac{2.88\times10^{-13}}{1.6\times10^{-13}}\ \mathrm{MeV} \\ &= 1.8\ \mathrm{MeV} \end{aligned} \)
From the above, \( U=1.8\ \mathrm{MeV} \). Hence, two protons 0.8 fm apart have an electrostatic potential energy of about 1.8 MeV.
NCERT (2026-27):
Summary – “6. … Potential energy of two charges q1, q2 at r1, r2 is given by”
\( \displaystyle U=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r_{12}} \)
“where r12 is distance between q1 and q2.”
Nuclear Force – “… The potential energy is a minimum at a distance r0 of about 0.8 fm. …”