Two protons are kept at a separation of 40 te | vedclass

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(C) The nuclear force $F_{n}$ is a short-range force that acts effectively only within the range of approximately $10^{-15} 	ext{ m}$ (or $1 	ext{ f

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$F_{n} \ll F_{e}$

Vedclass · Answer

(C) The nuclear force $F_{n}$ is a short-range force that acts effectively only within the range of approximately $10^{-15} 	ext{ m}$ (or $1 	ext{ fm}$). Beyond this distance,the nuclear force decreases exponentially and becomes negligible. The given separation is $40 	ext{ Å} = 40 	imes 10^{-10} 	ext{ m} = 4 	imes 10^{-9} 	ext{ m}$. Since $4 	imes 10^{-9} 	ext{ m} \gg 10^{-15} 	ext{ m}$,the nuclear force $F_{n}$ is essentially zero at this distance. However,the electrostatic force $F_{e}$ follows the inverse-square law $(F_{e} \propto 1/r^2)$ and remains significant even at this distance. Therefore,$F_{n} \ll F_{e}$.

Two protons are kept at a separation of $40 \text{ Å}$. $F_{n}$ is the nuclear force and $F_{e}$ is the electrostatic force between them. Then:

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