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Last updated date: 06th Dec 2023
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The unit of length convenient on the nuclear scale is a fermi : $1f = {10^{ - 15}}m$ . Nuclear sizes obey roughly the following empirical relation : $r = {r_0}{A^{1/3}}$ where, $r$ is the radius of the nucleus, $A$ its mass number, and ${r_0}$ is a constant equal to about $1.2f$ . Show that the rule implies that nuclear mass density is nearly constant for different nuclei. Estimate the mass density of sodium nucleus.

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Hint: Volume of the nucleus of radius $r$ is given by $V = \dfrac{4}{3}\pi {r^3}$ . Use the given expression $r = {r_0}{A^{1/3}}$ to find the value of radius of the nucleus.
Density of nucleus is given by, ${\text{Density}} = \dfrac{{{\text{Mass of nucleus}}}}{{{\text{Volume of nucleus}}}}$ . Here the mass of the nucleus will be in kg and not in amu.

We first use the given relation $r = {r_0}{A^{1/3}}$ in finding the volume of the nucleus.
As we know that the volume of the nucleus of radius $r$ is given by $V = \dfrac{4}{3}\pi {r^3}$
So, substituting the expression of radius in the formula of volume we have
$\implies$ $V = \dfrac{4}{3}\pi {\left( {{r_0}{A^{1/3}}} \right)^3} = \dfrac{4}{3}\pi A{r_0}^3$
Now, as we know that the mass of nucleus can be obtained through mass number after converting the unit amu to kg i.e. $M = A{\text{ amu}} = A \times 1.67 \times {10^{ - 27}}{\text{ kg}}$
Now, we know that the density of nucleus is given by $\rho = \dfrac{{{\text{Mass of nucleus}}}}{{{\text{Volume of nucleus}}}}$ . Here the mass of the nucleus will be in kg.
So, substituting the values in the above expression we have
$\implies$ $\rho = \dfrac{{A \times 1.67 \times {{10}^{ - 27}}}}{{\dfrac{4}{3}\pi A{r_0}^3}}$
As given in the question ${r_0} = 1.2f$ . So, substituting this in the expression we have
$\implies$ $\rho = \dfrac{{3 \times 1.67 \times {{10}^{ - 27}}}}{{4 \times 3.14 \times {{\left( {1.2 \times {{10}^{ - 15}}} \right)}^3}}} = \dfrac{{5.01 \times {{10}^{18}}}}{{21.71}} = 2.3 \times {10^{17}}{\text{ kg}}{{\text{m}}^{{\text{ - 3}}}}$
Hence, the mass density of sodium nucleus is $2.3 \times {10^{17}}{\text{ kg}}{{\text{m}}^{ - 3}}$ .

Note: In the expression for radius of nucleus $r = {r_0}{A^{1/3}}$ , ${r_0}$ is known as empirical constant. We can find in the expression of density of the nucleus that it is independent of the mass no. of the element or atom.