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An unknown nucleus has a nuclear density of $2.29 \times 10^{17}\ kg/m^3$ and mass of $19.926 \times 10^{-27}\ kg$. Its mass number A is approximately: (Take $R_0 = 1.2 \times 10^{-15}$ m; $4\pi = 12.56$)
A
12
B
19
C
20
D
16
Detailed Solution
Mass = Volume × Density: $M = \left(\frac{4}{3}\pi R^3\right)\rho$
Using $R = R_0 A^{1/3}$: $M = \left(\frac{4}{3}\pi R_0^3 A\right)\rho$
$\Rightarrow A = \frac{3M}{4\pi R_0^3 \rho}$
$A = \frac{3 \times 19.926 \times 10^{-27}}{12.56 \times (1.2 \times 10^{-15})^3 \times 2.29 \times 10^{17}} \approx 12$