Question

The radius of nucleus varies with mass number as \[{A^{1/n}}\]. The value of ‘n’ is:

Answer 1

Hint: The volume of the nucleus is directly proportional to the total number of the nucleons which suggests that all the nuclei have nearly the same density. Now as it is mentioned in the question that the radius of the nucleus varies with the mass number as \[{A^{1/n}}\], So, equate both the equations with each other and compare them to get the final value of 'n' for our result.

Complete step by step answer:
The radius of a nucleus (r) is directly proportional to the cube roots of its mass number represented by (A).
The most compelling evidence that nucleons are closely packed in a nucleus is that the radius of a nucleus, R, is found to be given approximately by \[R = {R_0}{(A)^{1/3}}\] where A is the mass number of the nucleus \[and,\;so,\;R \propto {A^{1/3}}\]
or, we can say that, the radius of a nucleus(R) is proportional to the cube root of its mass number(A).The volume of the nucleus is directly proportional to the total number of nucleons. This suggests that all nuclei have nearly the same density. Nucleons combine to form a nucleus as though they were tightly packed spheres. While the atomic no increases the electrons in the atom increases and no of shells increases so the size of the atom increases.
So, according to the law;
\[R = {R_0}{(A)^{1/3}}\];
Where, \[{R_0} = 1.2 \times {10^{ - 15}}m;\]
\[and,\;so,\;R \propto {A^{1/3}}\] …(1)
\[Given\;in\;question\;is;\;R \propto {A^{1/n}}\] …(2)
Equating equation (2) with equation (i);
\[ \Rightarrow \;R \propto {A^{1/n}} \propto {A^{1/3}}\]
\[ \Rightarrow \;\dfrac{1}{n} = \dfrac{1}{3}\]
\[ \Rightarrow \;n = 3\]
Therefore, the value of ‘n’ is 3 in this particular question.

Note:
According to nuclear theory, while the atomic number increases, the electrons in the atom increases and so the number of the shells increases and hence the size of the atom also increases.