Q147 / 180 · KCET 2008 · 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
    $\mathrm{F}_{\mathrm{n}}=\mathrm{F}_{\mathrm{e}}$
  • Correct answer
    C
    $\mathrm{F}_{\mathrm{n}} \ll \mathrm{F}_{\mathrm{e}}$
  • D
    $\mathrm{F}_{\mathrm{n}} \approx \mathrm{F}_{\mathrm{e}}$

Solution

$\mathrm{F}_{\mathrm{n}}$ 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 $40 Å \mathrm{F}_{\mathrm{n}} < \mathrm{F}_{\mathrm{e}}$.

Hence, the correct answer is option: 'C'.

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