Question

Calculate the radius of oxygen nucleus. Given that ${R_o} = 1.1 \times {10^{ - 16}}m$.

Answer 1

Hint: Nucleus is the centre of any atom which is surrounded by the electrons revolving the nucleus in their prescribed shells. Nucleus of an atom is very small as compared to the atom as a whole. The radius of the atom will be dependent on the atomic mass of the particular atom.

Complete answer:
We have a formula to calculate the radius of the nucleus of an atom by putting the values in the formula.
So, the formula for the radius of the oxygen nucleus is,

$R = {R_0}{A^{\left( {\dfrac{1}{3}} \right)}}$------equation (1) , where $A$ is the atomic mass and value of ${R_0}$ is given.
Given, ${R_o} = 1.1 \times {10^{ - 16}}m$
We know that the atomic mass of the oxygen, $A = 16$
So now putting the values in the equation (1), we get
$R = 1.1 \times {10^{ - 16}} \times {(16)^{\left( {\dfrac{1}{3}} \right)}}$
$ \Rightarrow R = 1.1 \times {10^{ - 16}} \times (2.52)$
$ \Rightarrow R = 27.72 \times {10^{ - 16}}m$
So, the radius of the oxygen nucleus is $27.72 \times {10^{ - 16}}m$.

Note:
The nucleus of the oxygen atom is having 8 protons and 8 neutrons. As we know that the number of the electrons is the same as that of the number of protons in an electrically neutral atom, then there are 8 electrons in the atom of the oxygen. These 8 electrons occupy the shells that are available around the nucleus. The chemical and physical properties will be determined by the stability of the outermost shell of the atom. Oxygen is vital for life on the earth. We human beings need this for respiration, that’s why this is a very important element for us. It is a mom-metal in the group 16, period 2, and the p-block of the periodic table. As a gaseous form the oxygen forms 20% of the atmosphere. It liquefies at the minus $219^\circ C$.