The density of a nucleus varies with the mass number $A$ as:

(A)^{2}

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(A)^{1 / 3}

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(A)^{0}

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(A)

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Solution

The correct option is C $(A)^{0}$
Atomic mass $=$ densityvolume of atom.
So, density of nucleus is in depend of mass number
$d = \frac{M a s s n o .}{\frac{4}{3} π r^{3}}$
$= \frac{A}{\frac{4}{3} π \times [1.3 \times 10^{- 13} \times A^{1 / 3}]^{3}} = c o n s t a n t A^{0}$
as $R = R_{0} A^{1 / 3}$ $(R_{0} = 1.1 \times 10^{- 15} m)$

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The density of a nucleus varies with the mass number A as:

The correct option is C (A)0Atomic mass = density*volume of atom.So, density of nucleus is in depend of mass number d=Massno.43πr3=A43π×[1.3×10−13×A1/3]3=constant∗A0as R=R0A1/3 (R0=1.1×10−15m)

The density of a nucleus varies with the mass number $A$ as: