The radius of R of a nucleus of mass number
Question:

The radius of $R$ of a nucleus of mass number $\mathrm{A}$ can be estimated by the formula $\mathrm{R}=\left(1.3 \times 10^{-15}\right) \mathrm{A}^{1 / 3} \mathrm{~m}$. It follows that the mass density of a nucleus is of the order of:

$\left(\mathrm{M}_{\text {port. }} \cong \mathrm{M}_{\text {neut. }} \simeq 1.67 \times 10^{-27} \mathrm{~kg}\right)$

  1. $10^{24} \mathrm{~kg} \mathrm{~m}^{-3}$

  2. $10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$

  3. $10^{17} \mathrm{~kg} \mathrm{~m}^{-3}$

  4. $10^{10} \mathrm{~kg} \mathrm{~m}^{-3}$


Correct Option: , 3

Solution:

$\rho_{\text {weloss }}=\frac{\text { mass }}{\text { volume }}=\frac{\mathrm{A}}{(4 / 3) \pi r_{0}^{3} \mathrm{~A}}=\frac{3}{4 \pi r_{0}^{3}}=2.3 \times 10^{17} \mathrm{~kg} / \mathrm{m}$

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