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Question: The unit of length convenient on the nuclear scale is a fermi : \(1f = {10^{ - 15}}m\) . Nuclear siz...

The unit of length convenient on the nuclear scale is a fermi : 1f=1015m1f = {10^{ - 15}}m . Nuclear sizes obey roughly the following empirical relation : r=r0A1/3r = {r_0}{A^{1/3}} where, rr is the radius of the nucleus, AA its mass number, and r0{r_0} is a constant equal to about 1.2f1.2f . Show that the rule implies that nuclear mass density is nearly constant for different nuclei. Estimate the mass density of sodium nucleus.

Explanation

Solution

Volume of the nucleus of radius rr is given by V=43πr3V = \dfrac{4}{3}\pi {r^3} . Use the given expression r=r0A1/3r = {r_0}{A^{1/3}} to find the value of radius of the nucleus.
Density of nucleus is given by, Density=Mass of nucleusVolume of nucleus{\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=r0A1/3r = {r_0}{A^{1/3}} in finding the volume of the nucleus.
As we know that the volume of the nucleus of radius rr is given by V=43πr3V = \dfrac{4}{3}\pi {r^3}
So, substituting the expression of radius in the formula of volume we have
    \implies V=43π(r0A1/3)3=43πAr03V = \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 amu=A×1.67×1027 kgM = A{\text{ amu}} = A \times 1.67 \times {10^{ - 27}}{\text{ kg}}
Now, we know that the density of nucleus is given by ρ=Mass of nucleusVolume of nucleus\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 ρ=A×1.67×102743πAr03\rho = \dfrac{{A \times 1.67 \times {{10}^{ - 27}}}}{{\dfrac{4}{3}\pi A{r_0}^3}}
As given in the question r0=1.2f{r_0} = 1.2f . So, substituting this in the expression we have
    \implies ρ=3×1.67×10274×3.14×(1.2×1015)3=5.01×101821.71=2.3×1017 kgm - 3\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 2.3×1017 kgm32.3 \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=r0A1/3r = {r_0}{A^{1/3}} , r0{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.