Question

Volume (V) of the nucleus is related to mass number (A) as
A. $V \propto {A^2}$
B. $V \propto {A^{\dfrac{1}{3}}}$
C. $V \propto {A^{\dfrac{2}{3}}}$
D. $V \propto A$

Answer 1

Hint: To find the required relation between volume (V) and mass number (A) we are assuming the shape of the nucleus is spherical of the radius (R) and mass number (A).
Complete step-by-step answer:
To express the required relation between volume (V) and mass number (A) we take the experimental formula for the radius of the nucleus:
Which says that:-
The nuclear density for a nucleus can be approximately calculated from the size of the nucleus: Density${\text{(n) = }}\dfrac{{{\text{mass number}}}}{{{\text{size of nucleus}}}}$…………….(i)
Now,
Simplifying the Eqn (i), we have
Density${\text{(n) = }}\dfrac{{{\text{mass number}}}}{{{\text{size of nucleus}}}}$
Size of nucleus = Density$ \times $Mass number
Size of nucleus =$\dfrac{4}{3}\pi {R^3}$
$ \Rightarrow {R^3} = \dfrac{3}{4}\pi \times n \times A$
$ \Rightarrow R = {\left[ {\dfrac{3}{4}\pi \times n \times A} \right]^{\dfrac{1}{3}}}$
$ \Rightarrow R = {\left[ {\dfrac{3}{4}\pi \times n} \right]^{\dfrac{1}{3}}}{A^{\dfrac{1}{3}}}$
$ \Rightarrow R = {R_0}{A^{\dfrac{1}{3}}}$
So, we finally get:
$R = {R_0}{A^{\dfrac{1}{3}}}$……......... (ii), where [${R_0} = 1.1 \times {10^{ - 15}}{\text{m}}$${\text{ = 1}}{\text{.1 fm\;}}$ and A is the mass number, i.e. the sum of neutrons and protons in the nucleus.]

Calculating the volume of the spherical nucleus.
$V = \dfrac{4}{3}\pi {R^3}$…………..(iii)
Substituting the given value of (R) from equation (ii) into equation (iii).we get $V = \dfrac{4}{3}\pi {\left( {{R_0}{A^{\dfrac{1}{3}}}} \right)^3}$
$ \Rightarrow V = \dfrac{4}{3}\pi R_0^3A$
$ \Rightarrow V = kA$Where we take $\left[ {k = \dfrac{4}{3}\pi R_0^3} \right]$
$ \Rightarrow V \propto A$
Hence, the correct answer option is (D)$V \propto A$.

Note:One should note that the obtained relation by the assumption of the shape of the nucleus is spherical while it has a great advantage and a very deep concept in nuclear physics in higher study. So, for up to the higher secondary level of nuclear physics we simply use the spherical shape consideration for solving various numerical problems.