Q. Mass number $A $ of a nucleus whose radius is $3.9$ fermi is

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A

10

B

39

C

17

D

27

Solution:

$R=R_{o} A^{\frac{1}{3}}$
$3.9 \times 10^{-15}$
$=1.3 \times 10^{-15} A ^{1 / 3}$
$A^{1 / 3}=3$
therefore $A= 3 ^{3}= 2 7$

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