Question

If the nuclear radius of $ ^{27}A1 $ is $ 3.6Fermi $ , the approximate nuclear radius of $ ^{64}Cu $ in Fermi is,
(A) $ 4.8 $
(B) $ 3.6 $
(C) $ 2.4 $
(D) $ 1.2 $

Answer 1

Hint :Use the relation between the nuclear radius and the mass number of an element to find its nuclear radius. The relation between the nuclear radius and the mass number of an element is given by, $ r \propto {A^{1/3}} $ where, $ r $ is the nuclear radius and $ A $ is the mass number of the element. ( Mass number = Neutron Proton Number)

Complete Step By Step Answer:
We know that the relation between the nuclear radius and the mass number of an element is given by, $ r \propto {A^{1/3}} $ where, $ r $ is the nuclear radius and $ A $ is the mass number of the element. ( Mass number = Neutron Proton Number)
So, equating the relation we get, $ r = {r_0}{A^{1/3}} $ where, $ {r_0} $ is a constant the value of which is the same for all the elements. Now we have given here aluminium with mass number $ 27 $ and its nuclear radius is $ 3.6Fermi $ . Putting these values in the equation we get,
 $ r = {r_0}{\left( {27} \right)^{1/3}} $
Or, $ 3.6 = 3{r_0} $
Therefore we get the value of $ {r_0} $ as,
 $ {r_0} = \dfrac{{3.6}}{3} = 1.2fm $
Now, we have to find the nuclear radius of copper ( $ ^{64}Cu $ ), we have given the mass number of copper is $ A = 64 $ .Hence, putting the value of the constant $ {r_0} = 1.2fm $ and mass number $ A = 64 $ in the equation we get,
 $ r = {r_0}{A^{1/3}} $
Or, $ r = 1.2{\left( {64} \right)^{1/3}} $
On simplifying that becomes,
 $ r = 4.8fm $
Therefore, the nuclear radius of copper with mass number $ A = 64 $ is $ 4.8fm $ .
Hence, Option ( A ) is the correct option.

Note :
From the relation we can see that the nuclear radius increases as mass number increases, the nuclear radius increases. But, we know that when more and proton and neutron are there then attraction between them is also more high, So, at large values of $ A $ the nuclear radius does not increase by that much. At large values of $ A $ it increases slightly. If we look at the variation of the equation $ r \propto {A^{1/3}} $ we can realize that easily.