The density of a nucleus in which mass of each nucleon is $1.67 \times 10^{- 27} kg$ and $R_{0} = 1.4 \times 10^{- 15} m$ is

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$1.453 \times 10^{17} kg / m^{3}$

$1.453 \times 10^{10} kg / m^{3}$

$1.453 \times 10^{16} kg / m^{3}$

$1.453 \times 10^{21} kg / m^{3}$

answer is A.

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Detailed Solution

$d = \frac{Am}{\frac{4}{3} {πR}^{3}} = \frac{3 Am}{4 π {(R_{0} A^{\frac{1}{3}})}^{3}} = \frac{3 Am}{4 {πR}_{0}^{3} A} = \frac{3 m}{4 {πR}_{0}^{3}}$

$\Rightarrow d = \frac{3 \times 1.67 \times 10^{- 27}}{4 π \times {(1.4 \times 10^{- 15})}^{3}} = 1.453 \times 10^{17} kg / m^{3}$

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