The radius R of a nucleus of mass number A can be estimated by the formula $R = (1.3 \times 10^{- 15}) A^{1 / 3} m$ . It follows that the mass density of a nucleus is of the order of : $(M_{prot} ≅ M_{neut} = 1.67 \times 10^{- 27} kg)$

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$10^{17} k g m^{- 3}$

$10^{24} k g m^{- 3}$

$10^{10} k g m^{- 3}$

$10^{3} k g m^{- 3}$

answer is A.

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Detailed Solution

$d e n s i t y = \frac{M_{P}}{V_{P}}$

$\begin{array}{l} = \frac{1.67 \times 10^{- 27} \times A}{\frac{4}{3} π {(1.3 \times 10^{- 15})}^{3} (A)} kg / m^{3} \\ = \frac{3 \times 1.67}{4 π \times (1.3)^{3}} \times 10^{18} kg / m^{3} \\ = \frac{30 \times 1.67}{4 π (1.3)^{3}} \times 10^{17} kg / m^{3} \\ ∴ d = 1.81 \times 10^{17} kg / m^{3} \end{array}$