The Coulomb barrier for two protons is about 400 keV. The separation at which the electrostatic potential energy of two protons is 400 keV is about
(Take \(\frac{1}{4\pi\varepsilon_0} = 9\times10^{9}\ \mathrm{N\,m^{2}\,C^{-2}}\) and \(e = 1.6\times10^{-19}\ \mathrm{C}\))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(3.6\ \mathrm{fm}\)
The potential energy of two charges \( e \) at a distance \( r \) is \( U=\frac{1}{4\pi\varepsilon_0}\frac{e^{2}}{r} \). First change the energy into joule:
\( \begin{aligned} U &= 400\times10^{3}\times1.6\times10^{-19}\ \mathrm{J} \\ &= 6.4\times10^{-14}\ \mathrm{J} \end{aligned} \)
Now find the separation:
\( \begin{aligned} r &= \frac{1}{4\pi\varepsilon_0}\,\frac{e^{2}}{U} \\ &= \frac{9\times10^{9}\times(1.6\times10^{-19})^{2}}{6.4\times10^{-14}} \\ &= 3.6\times10^{-15}\ \mathrm{m} \\ &= 3.6\ \mathrm{fm} \end{aligned} \)
Therefore, two protons must come within about 3.6 fm of each other before their electrostatic potential energy reaches 400 keV.
NCERT (2026-27):
Potential Energy of a System of Charges – “… Thus, the potential energy of a system of two charges q1 and q2 is”
\( \displaystyle U=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r_{12}} \)\( \qquad (2.22) \)