Mass number of He is 4 and that for sulphur is 32. radius of sulphur nucleus is larger than that of

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13.Nuclei

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The mass number of $He$ is $4$ and that for sulphur is $32.$ The radius of sulphur nucleus is larger than that of helium, by times

A

$\sqrt 8 $

B

$4$

C

$2$

D

$8$

(AIPMT-1995)

Solution

(c) $r \propto {(A)^{1/3}}$

$\frac{R_{S}}{R_{H e}}=\left(\frac{A_{S}}{A_{H e}}\right)^{1 / 3}=\left(\frac{32}{4}\right)^{1 / 3}=2$

Standard 12

Physics

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