Question

The nuclear radius of a certain nucleus is \[7.2fm\] and it has a charge of \[1.28 \times {10^{ - 17}}C\]. The number of neutrons inside the nucleus is
$\left( {\text{A}} \right)$\[136\]
$\left( {\text{B}} \right)$$140$
$\left( {\text{C}} \right)$$126$
$\left( {\text{D}} \right)$$142$

Answer 1

Hint:Use the equation of relation between the radius of a nucleus, the experimental value of the radius of a proton or neutron, and the atomic mass(=the total number of protons and neutrons in the nucleus).
From the equation, by putting the given values we can get the value of atomic mass.
Calculate the number of the proton from the given charge of the nucleus since neutrons are neutral charged particles of the nucleus. And hence find the number of the neutron by subtracting the proton numbers from the atomic mass or the total number of the protons and neutrons in the nucleus.

Formula used:
\[R = {R_0}{A^{\dfrac{1}{3}}}\]
Where $R$= The radius of the nucleus,
$A$ is equal to the total number of protons and neutrons in the nucleus,
${R_0}$= the radius of a proton or a neutron.
The number of protons, ${\text{Z = }}\dfrac{{{\text{charge of the nucleus}}}}{{{\text{charge of a proton}}}}$.
The number of neutrons = $A - Z$.

Complete step by step answer:
The radius of the nucleus is$R$, the atomic mass=the total number of protons and neutrons in the nucleus=$A$ and the radius of a proton or a neutron is ${R_0}$.
So, the volume of the nucleus, $V = \dfrac{4}{3}\pi R_{}^3$
And, the volume of a proton or a neutron, $V' = \dfrac{4}{3}\pi R_0^3$
So, $V = V' \times A$

$ \Rightarrow \dfrac{4}{3}\pi R_0^3 = \dfrac{4}{3}\pi R_{}^3 \times A$
$ \Rightarrow R_0^3 = {R^3}A$
\[ \Rightarrow A = {\left( {\dfrac{R}{{{R_0}}}} \right)^3}\]
Since, the experimental value of the radius of a proton or a neutron is ${R_0} = 1.2fm$
\[ \Rightarrow A = {\left( {\dfrac{{7.2}}{{1.2}}} \right)^3}\]
\[ \Rightarrow A = {6^3}\]
\[ \Rightarrow A = 216\]
The number of proton in the nucleus,
${\text{Z = }}\dfrac{{{\text{charge of the nucleus}}}}{{{\text{charge of a proton}}}}$
\[Z = \dfrac{{1.28 \times {{10}^{ - 17}}}}{{1.6 \times {{10}^{ - 19}}}}\]
\[ \Rightarrow Z = 0.8 \times 100\]
\[ \Rightarrow Z = 80\]
The number of neutrons in the nucleus,
\[A - Z = 216 - 80\]
\[ \Rightarrow A - Z = 136\]

Hence the right answer is in option $\left( {\text{A}} \right)$. \[\]

Notes:
From the experimental value of the radius of a proton or a neutron we can get the nuclear density by the following steps,
Nuclear density, ${\rho _N} = \dfrac{M}{V}$, where $M = $ $Au = A \times 1.66 \times {10^{ - 24}}gm$
$\therefore {\rho _N} = \dfrac{{A \times 1.66 \times {{10}^{ - 24}}}}{{\dfrac{4}{3}\pi R_0^3A}}$, where ${R_0} = 1.2fm = 1.2 \times {10^{ - 13}}cm$
$ \Rightarrow {\rho _N} = \dfrac{{1.66 \times {{10}^{ - 24}}}}{{\dfrac{4}{3}\pi {{(1.2 \times {{10}^{ - 13}})}^3}}}$
On simplification we get,
$ \Rightarrow {\rho _N} = 2.3 \times {10^{14}}g/c{m^3}$
This value implies that the density of a nucleus is very high and hence a large number of objects remain concentrated in a very small space.