Given: r = R₀A^⅓, where r is the radius of nucleus of mass no. A and R₀ is a constant whose value is equal to 1.5 × 10^1-15metre. (Given 1 amu = 1.66 × 10^-15 g)
What is the density of a nucleus of mass number A?

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$1.17 \times 10^{17} kg / {cm}^{3}$

$1.17 \times 10^{- 17} kg / m^{3}$

$\frac{4}{3} π {(1 .5 \times 10^{- 15})}^{3} A$

None of these

answer is B.

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Detailed Solution

$r = R_{0} A^{1 / 3}$

$∴ Volume of nucles (4 / 3) π r^{3}$

$\begin{array}{l} = \frac{4}{3} π {(1 .5 \times 10^{- 15})}^{3} m^{3} \\ density d = \frac{m}{V} = \frac{1 .66 \times 10^{- 27} kg}{4 / 3 \times 3 .14 {(1 .5 \times 10^{- 15})}^{3}} \\ = 1 .17 \times 10^{17} kg / m^{3} \end{array}$

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