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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.

Answer
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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.

Complete step by step answer:
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 $2.3 \times {10^{17}}{\text{ kg}}{{\text{m}}^{ - 3}}$ .

Additional Information:
There is a device called Nuclear Density Gauge (NZO) which is used to find the density of several materials that can be further used in the construction of roads, highways and airports runways. This device works with the help of gamma radiations and uses its interaction with materials to find its density. These devices are also used in the petroleum industry.


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.
Last updated date: 28th May 2023
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