Question

The stable nucleus that has a radius ${\raise0.7ex\hbox{$1$} \!\mathord{\left/
 {\vphantom {1 3}}\right.}\!\lower0.7ex\hbox{$3$}}$ that $O{s^{189}}$ is
(A) $_3L{i^7}$
(B) $_2H{e^4}$
(C) $_5{B^{10}}$
(D) $_6{C^{12}}$

Answer 1

Hint: The radius and mass number of any nucleus are related by the formula, $R = 1.2 \times {10^{ - 15}}{A^{{1 \mathord{\left/
 {\vphantom {1 3}} \right.} 3}}}$. So we can find the radius of the nucleus of $O{s^{189}}$ and from there we can find the radius of the other stable nuclei as it is one-third of the radius of $O{s^{189}}$ . From there by again using the same formula we can find the mass number of the stable nuclei.

Formula used: In the solution to this question, we will be using the following formula,
$\Rightarrow R = 1.2 \times {10^{ - 15}}{A^{{1 \mathord{\left/
 {\vphantom {1 3}} \right.} 3}}}$
where $R$ is the radius of the nuclei and $A$ is the mass number of the same nuclei.

Complete step by step answer:
For any stable nuclei, the mass number and the radius are related by the formula,
$\Rightarrow R = 1.2 \times {10^{ - 15}}{A^{{1 \mathord{\left/
 {\vphantom {1 3}} \right.} 3}}}$
So for the given species, $O{s^{189}}$ we have the mass number given as 189. So by substituting that value in the above formula we can calculate the radius of $O{s^{189}}$.
Therefore, $R = 1.2 \times {10^{ - 15}}{\left( {189} \right)^{{1 \mathord{\left/
 {\vphantom {1 3}} \right.} 3}}}$
We keep the value of the radius in this form. Now it is given in the question that a stable nucleus has a radius one-third of the radius of $O{s^{189}}$.
So if the radius of the stable nucleus be ${R_1}$ then the value of ${R_1}$ is given by,
$\Rightarrow {R_1} = \dfrac{R}{3}$
Substituting the value of $R$ we get,
$\Rightarrow {R_1} = \dfrac{{1.2 \times {{10}^{ - 15}}{{\left( {189} \right)}^{{1 \mathord{\left/
 {\vphantom {1 3}} \right.} 3}}}}}{3}$
Now again let the mass number of this stable nucleus be ${A_1}$. So the radius and the mass number of this nucleus are again related as,
$\Rightarrow {R_1} = 1.2 \times {10^{ - 15}}{A_1}^{{1 \mathord{\left/
 {\vphantom {1 3}} \right.} 3}}$
So by equating the two values of ${R_1}$ we get,
$\Rightarrow 1.2 \times {10^{ - 15}}{A_1}^{{1 \mathord{\left/
 {\vphantom {1 3}} \right.} 3}} = \dfrac{{1.2 \times {{10}^{ - 15}}{{\left( {189} \right)}^{{1 \mathord{\left/
 {\vphantom {1 3}} \right.} 3}}}}}{3}$
From the above equation by cancelling $1.2 \times {10^{ - 15}}$ from the numerator of both the sides we get,
$\Rightarrow {A_1}^{{1 \mathord{\left/
 {\vphantom {1 3}} \right.} 3}} = \dfrac{{{{\left( {189} \right)}^{{1 \mathord{\left/
 {\vphantom {1 3}} \right.} 3}}}}}{3}$
Therefore to get the value of ${A_1}$ we apply cube to both the sides and get,
$\Rightarrow {A_1} = \dfrac{{189}}{{{3^3}}} = \dfrac{{189}}{{27}}$
On calculation this gives,
$\Rightarrow {A_1} = 7$
So the mass number of the stable nucleus is ${A_1} = 7$.
In the given options, the nucleus $_3L{i^7}$ has a mass number of 7.
So the correct option is (A); $_3L{i^7}$.

Note:
Since the spherical volume of a nucleus increases with the increase in the number of the nucleons so the volume is directly proportional to the mass number. In the volume the radius varies with the power cube, so ${R^3} \propto A$. By applying cube roots on both sides, we have $R \propto {A^{{1 \mathord{\left/{\vphantom {1 3}} \right.} 3}}}$. On removing the constant of proportionality, we get a constant whose value is equal to $1.2 \times {10^{ - 15}}$.