Question

The radius of a spherical nucleus as measured by electron scattering is $ 3.6fm $ . What is the likely mass number of the nucleus?
 $ \left( A \right)27 \\
  \left( B \right)40 \\
  \left( C \right)56 \\
  \left( D \right)120 \\ $

Answer 1

Hint : In order to solve this question, we are going to use the radius of the spherical nucleus as given in the question and the radius $ {r_0} $ , the radius of a spherical nucleus is directly proportional to the mass number raised to the power $ \dfrac{1}{3} $ , this relation gives us a simpler relation to evaluate the value of mass number.
The radius of a nucleus is given by the formula
 $ r = {r_0}{A^{\dfrac{1}{3}}} $
Where $ r $ is the radius of the spherical nucleus as measured by the electron scattering and $ {r_0} $ is the Bohr’s radius.

Complete Step By Step Answer:
As given in the question, the radius of the spherical nucleus as measured by the electron scattering is $ 3.6fm $ , and we know that the value of $ {r_0} $ is $ 1.2fm $
Now, the radius of a nucleus is given by the formula
 $ r = {r_0}{A^{\dfrac{1}{3}}} $
Solving this, we get the mass number relation to be
 $ \Rightarrow A = {\left( {\dfrac{r}{{{r_0}}}} \right)^3} $
Putting the two values in this, we get
 $ \Rightarrow A = {\left( {\dfrac{{3.6fm}}{{1.2fm}}} \right)^3} = 27 $
Thus, option $ \left( A \right)27 $ is the correct answer.

Note :
The Bohr radius $ {r_0} $ is equal to the most probable distance between the nucleus and the electron in an atom, where $ r $ is the radius of the atom. Here we first use the radius formula to correlate the relationship between r and A, then we put the value of this to A.
We have to clarify first that both the radii units should be the same so as to get a numerical value.