Question

Approximate size of the nucleus of $_{28}^{64}Ni$ is

A. 3.5 fm
B. 4.5 fm
C. 5.5 fm
D. 2.5 fm

Answer 1

Hint: To solve the question we have to use the formula of nuclear size which is given as $r={{R}_{0}}{{A}^{\dfrac{1}{3}}}$ where, r represents the radius of nucleus, ${{R}_{0}}$ is the constant value which is equal to 1.25 fm and A represents the atomic mass of the element. Atomic mass of Ni is given as 64 and atomic number of Ni is given as 28.

Complete step by step solution:
From your chemistry lessons you have learned about Rutherford's experiment and how to find the size of the nucleus. Rutherford performed an experiment to estimate the size of the nucleus, the experiment is done by scattering alpha particles from the nucleus which is considered as the point charge particle. To find the size of the nucleus we have to obtain the point of closest approach of the alpha particle. So, when we shoot the alpha particles of kinetic energy 5.5MeV then the point of closest approach was estimated to be $4\times {{10}^{-4}}m$. And as we know that repulsive force that will act here is coulomb repulsion so there will be no contact. Hence, we can say that the size of the nucleus will be smaller than $4\times {{10}^{-4}}m$. Now, to determine the accurate measure of the size of the nucleus of various elements many more iterations of experiment have been carried out. As a result a formula to measure the size of the nucleus was determined. The formula is given as:
\[r={{R}_{0}}{{A}^{\dfrac{1}{3}}}\]
Where, r = Radius of the nucleus
${{R}_{0}}$ = constant value which is equals to 1.25 fm
and A = Atomic mass of the element
So, in the question the value of atomic mass of the Ni is given as 64 and the value of atomic number is given as 28. Now put the value of atomic mass in the formula, we will get:
\[r=1.25\times {{(64)}^{\dfrac{1}{3}}}\]
\[r=1.24\times {{(4)}^{3\times \dfrac{1}{3}}}\]
\[\therefore r=1.25\times 4=5fm\]
Hence the size of the nucleus is 5 fm.

Thus the correct option will be (C).

Note: From the formula we can say that the volume of the nucleus which is proportional to ${{R}^{3}}$ will be proportional to A (atomic mass) and as we see that there is no mention of density in the formula because density of the nucleus does not vary according to the elements. It is approximately $2.3\times {{10}^{17}}Kg.{{m}^{-3}}$. fm represents the fermi units which is equals to ${{10}^{-15}}m$.