Two \({}^{3}_{2}\mathrm{He}\) nuclei move towards each other head on. Take each nucleus as a hard sphere of radius 1.8 fm. The height of the Coulomb barrier, when the two nuclei just touch, is about
(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 2
The correct answer is Option (2) → \(1.6\ \mathrm{MeV}\)
The barrier height is the electrostatic potential energy of the two nuclei when they just touch. Each \( {}^{3}_{2}\mathrm{He} \) nucleus has the charge \( 2e \). When the spheres touch, their centres are \( 2\times1.8 = 3.6\ \mathrm{fm} \) apart.
\( \begin{aligned} U &= \frac{1}{4\pi\varepsilon_0}\,\frac{(2e)(2e)}{r} \\ &= \frac{9\times10^{9}\times4\times(1.6\times10^{-19})^{2}}{3.6\times10^{-15}} \\ &= 2.56\times10^{-13}\ \mathrm{J} \end{aligned} \)
In MeV:
\( \begin{aligned} U &= \frac{2.56\times10^{-13}}{1.6\times10^{-13}}\ \mathrm{MeV} \\ &= 1.6\ \mathrm{MeV} \end{aligned} \)
NCERT (2026-27):
Potential Energy of a System of Charges – “… Thus, the potential energy of a system of two charges q1 and q2 is”
\( \displaystyle U=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r_{12}} \)\( \qquad (2.22) \)