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Two protons are kept at a separation of $40 Å F_{n}$ is the nuclear force and $F_{\varepsilon}$ is the electrostatic force between them. Then,
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The correct answer is:
$F_{n} \gg>F_{e}$
Nuclear force $F_{n}$ are attractive in nature. As in the given case, the two protons are kept at a separation of $40 Å$ (short range), hence in this range, nuclear force $F_{n}$ is much greater than the electrostatic force $F_{e}$.
i.e. $F_{n} \gg F_{e}$
i.e. $F_{n} \gg F_{e}$
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