KCET · Physics · Nuclear Physics
Two protons are kept at a separation of \(40 Å . F_{n}\) is the nuclear force and \(\mathrm{F}_{\mathrm{e}}\) is the electrostatic force between them. Then
- A \(\mathrm{F}_{\mathrm{n}} \gg \mathrm{F}_{\mathrm{e}}\)
- B \(\mathrm{F}_{\mathrm{n}}=\mathrm{F}_{\mathrm{e}}\)
- C \(\mathrm{F}_{\mathrm{n}} \ll \mathrm{F}_{\mathrm{e}}\)
- D \(\mathrm{F}_{\mathrm{n}} \approx \mathrm{F}_{\mathrm{e}}\)
Answer & Solution
Correct Answer
(C) \(\mathrm{F}_{\mathrm{n}} \ll \mathrm{F}_{\mathrm{e}}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{F}_{\mathrm{n}}\) is stronger than \(\mathrm{F}_{\mathrm{e}} . \mathrm{F}_{\mathrm{n}}\) operates at very short range inside the nucleus as little as \(10^{-15} \mathrm{~m}\). As in the given case two protons are kept at a separation of \(40 Å \mathrm{F}_{\mathrm{n}} < \mathrm{F}_{\mathrm{e}}\).
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