Question

Considering the shape of $Ca$ nucleus like a sphere, calculate the density of the nucleus of $Ca$ of mass number 40.

Answer 1

In this question, we need to determine the density of the nucleus of $Ca$ of mass number 40. For this we will the relation between the volume of the nucleus and the radius along with the atomic number.

Complete answer:
Calcium is a typical heavy alkaline earth metal which has higher density and it is harder to be cut with a knife, it requires a lot of effort. Naturally calcium has five isotopes that exist. Previously, calcium compounds were known as millennia. Calcium lies in group 2 and period 4 of the periodic table.
Radius of nucleus of $Ca$ atom can be calculated by using formula, $R = {R_0}{A^{\frac{1}{3}}}$ (this formula to calculate the radius of nucleus of an atom in the periodic table) where, $A = 40$ for $Ca$ and value of ${R_0} = 1.2 \times {10^{ - 15}}m$ is the constant used to find out the radius of nucleus of an atom.
Substituting the values in the formula $R = {R_0}{A^{\frac{1}{3}}}$, we get $R = 1.2 \times {10^{ - 15}} \times {\left( {40} \right)^{\left( {\dfrac{1}{3}} \right)}}$
The above value gives us the value of the radius of the nucleus of $Ca$ atom.
Now, the volume of the nucleus can be calculated by using the equation $V = \dfrac{4}{3}\pi {R^3}$ as the atom is of spherical shape.
So, substituting the values in the equation $V = \dfrac{4}{3}\pi {R^3}$ to calculate the value of volume occupied by the spherical atom.

$$V = \dfrac{{4 \times \pi \times {{\left( {1.2 \times {{10}^{ - 15}} \times {{(40)}^{\dfrac{1}{3}}}} \right)}^3}}}{3} \\\ = \dfrac{{4 \times 3.14 \times {{\left( {1.2} \right)}^3} \times {{10}^{ - 45}} \times 40}}{3} \\\ = \dfrac{{4 \times 3.14 \times 1.728 \times {{10}^{ - 45}} \times 40}}{3} \\\ = 289.53 \times {10^{ - 45}}{\text{ }}{{\text{m}}^3} \\\$$

Hence, the volume occupied by the nucleus of the Calcium atom is $289.53 \times {10^{ - 45}}{\text{ }}{{\text{m}}^3}$.
Now, mass of the nucleons, $M = 40\,amu$
And,
$1\,amu$ is $\dfrac{1}{{12}}$ of the mass of one carbon-$12$ atom.
It is equal to, $\dfrac{1}{{{N_A}}}$ $= \dfrac{1}{{6.022 \times {{10}^{23}}}}$
So, $1\,amu = 1.67 \times {10^{ - 27}}kg$
So, Mass of nucleons in $kg$ $= 40 \times 1.67 \times {10^{ - 27}}\,kg$
Mass of nucleons means mass of protons and neutrons present in the nucleus of an atom.
And then,
Using formula, $density = \dfrac{{mass}}{{volume}}$
Putting the value of mass and volume in the above equation gives us the value of density of $Ca$ atom.
$\Rightarrow density = \dfrac{{40 \times 1.67 \times {{10}^{ - 27}}}}{{289.53 \times {{10}^{ - 45}}}}kg{m^{ - 3}}$
Calculating it gives the value of density of $Ca$ atom.
$\Rightarrow density = 0.23 \times {10^{18}}kg{m^{ - 3}}$
This is the density of $Ca$ atom having mass of nucleons as 40.

Note:
$Ca$ is considered as a sphere that is why formula to calculate volume of sphere atom $V = \dfrac{{4\pi {R^3}}}{3}$is used, this is kept in mind. Volume of nucleus of $Ca$ is $V = \dfrac{{4\pi {R^3}}}{3}$ where, $R = {R_0}{A^{\frac{1}{3}}}$, $A$ is the mass number.