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.