Question

Two nuclei have mass numbers in the ratio \[1:8\], what is the ratio of their nuclear radii?

Answer 1

Hint: We must know the relationship between mass number and radius of nuclei to solve this question. Mass number of a nucleus is the number of nucleons present in a nucleus of an atom. The radius of the nucleus is directly proportional to the cube root of mass number. So we will use this relation to find the ratio of the nuclear radii of the given two nuclei.

Formula used:
\[R={{R}_{0}}{{\left( A \right)}^{\dfrac{1}{3}}}\]

Complete step by step answer:
We know, the mass number\[A\]and the radius of atomic nuclei is related by the expression,
    \[R={{R}_{0}}{{\left( A \right)}^{\dfrac{1}{3}}}\]
Where, \[R\] is the radius of the nucleus.
               \[A\] is the mass number
And, \[{{R}_{0}}\] is a constant known as fermi constant which is same for all the nucleus and have a value of \[1.2\times {{10}^{-15}}m\].
Now, the ratio of mass number is given as \[1:8\]. So, let us assume the mass number of the first nucleus to be \[A\] and the second nucleus be \[8A\].
Then the radius of the first nucleus will be, \[{{R}_{1}}={{R}_{0}}{{\left( A \right)}^{\dfrac{1}{3}}}\].
And the radius of the second nucleus will be, \[{{R}_{2}}={{R}_{0}}{{\left( 8A \right)}^{\dfrac{1}{3}}}\].
Now, if we take the ratio between these two radii, it will be
\[\dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{{{R}_{0}}{{\left( A \right)}^{\dfrac{1}{3}}}}{{{R}_{0}}{{\left( 8A \right)}^{\dfrac{1}{3}}}}=\dfrac{1}{{{\left( 8 \right)}^{\dfrac{1}{3}}}}=\dfrac{1}{2}\]
Therefore, the ratio of the radii of two nuclei having a mass number ratio of\[1:8\]is found to be\[1:2\].

Note:
We must know that the nucleons constitute the mass number of a nucleus. Nucleons are particles present in atomic nuclei and we know nucleus is composed of protons and neutrons. Both of these particles' masses are equal, which is \[1.67\times {{10}^{-27}}kg\]. Protons are positively charged and neutrons have no charge. The fermi constant that we used will not change for any nucleus and it has a value of \[1.2fermi=1.2\times {{10}^{-15}}m\].