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Radius of ${}_{2}^{4}{He}$ nucleus is 3 Fermi. The radius of ${}_{82}^{206}{Pb}$ nucleus will be

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5 Fermi

6 Fermi

11.6 Fermi

8 Fermi

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Correct option is C

We have, r∝A1/3⇒r2r1=A2A11/3=20641/3∴r2=320641/3=11.6 Fermi.

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Radius of He 24 nucleus is 3 Fermi. The radius of Pb 82206 nucleus will be

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Radius of ${}_{2}^{4}{He}$ nucleus is 3 Fermi. The radius of ${}_{82}^{206}{Pb}$ nucleus will be