The radius of a nucleus with mass number $A = 64$ is approximately ($R_0 = 1.2$ fm):
A$77$ fm, multiplying the mass number directly here
B$4.8$ fm, since $R = R_0 A^{1/3} = 1.2\cdot 4$
C$1.2$ fm, equal to the constant alone for all nuclei
D$76.8$ fm, ignoring the cube-root in the formula
Answer & Solution
Correct answer: B. $4.8$ fm, since $R = R_0 A^{1/3} = 1.2\cdot 4$
$A^{1/3} = 4$, $R = 1.2\cdot 4 = 4.8$ fm.
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