Question

The ratio of the masses of the elements having the nuclear radii 2 fermi and 1 fermi is:
A. 8
B. 2
C. 3
D. 4

Answer 1

Hint: Radius of a nucleus is directly proportional to the cube root of its mass number. Therefore one can write the ratio for the two particles easily and on cubing the two sides, we get the ratio of the masses.

Formula used:
The radius of a nucleus of mass number A is written as:
$R = R_0 A^{1/3}$
where $R_0$ is known as Bohr radius and is a constant.

Complete step-by-step answer:
The radius of the first nucleus is:
$R_1 = R_0 A^{1/3}_1$
and of second nucleus is:
$R_2 = R_0 A^{1/3}_2$.

Taking the ratio, we get:
$\dfrac{R_1}{R_2} = \left( \dfrac{A_1}{A_2} \right)^{1/3}$.
Cubing both sides we get:
$ \left( \dfrac{R_1}{R_2} \right)^3 = \dfrac{A_1}{A_2} $ ,

keeping $R_1$ = 2 fermi and $R_2$ = 1 fermi, we get:
$\dfrac{A_1}{A_2} = 2^3 = 8$.

So, the correct answer is “Option A”.

Additional Information: Nucleus like a liquid is incompressible in nature and for a spherical liquid we may write:
$M = \rho \left( \dfrac{4}{3} \pi R^3 \right)$
where M is the mass of the spherical liquid, which is equal to density times volume.
Compare this with the formula that we used for the question where
$A \propto R^3$ .
Therefore by considering the fact that mass is directly proportional to volume, we can formulate the relation that we used here for solving our question.
Also, R being nuclear radius is the distance from the centre of the nucleus where a drop of 50% is seen in the density. This means that the nucleus is not a defined sphere of radius R, it is assumed to be so.

Note: The relation for radius has 'mass number' A instead of 'mass'. Mass number A is simply the sum of total number of neutrons and total number of protons in a nucleus. A is not the mass of a nucleus in kg or g but it happens to be the mass of the nucleus in u i.e., unified mass units.