Two protons are at the opposite ends of a diameter of a nucleus of radius 4.8 fm. The Coulomb force between the two protons 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 3
The correct answer is Option (3) → \(2.5\ \mathrm{N}\)
The two protons are one diameter, \( 2R \), apart. So the separation is \( r = 2R = 9.6\ \mathrm{fm} = 9.6\times10^{-15}\ \mathrm{m} \). Now use Coulomb’s law:
\( \begin{aligned} F &= \frac{1}{4\pi\varepsilon_0}\,\frac{e^{2}}{r^{2}} \\ &= \frac{9\times10^{9}\times(1.6\times10^{-19})^{2}}{(9.6\times10^{-15})^{2}} \\ &= \frac{2.304\times10^{-28}}{9.216\times10^{-29}} \\ &= 2.5\ \mathrm{N} \end{aligned} \)
From the above, the two protons repel each other with a force of about 2.5 N. At 9.6 fm the nuclear force between them is practically zero. So each of them is held in the nucleus by its near neighbours.
NCERT (2026-27):
Coulomb’s Law – “… so that Coulomb’s law is written as”
\( \displaystyle F=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^{2}} \)\( \qquad (1.2) \)