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Q.
The nucleus which has radius one-third of the radius of $Os ^{189}$ is
Nuclei
Solution:
Let nucleus be $q^{X^{A}}$.
Nuclear radius, $R=R_{0}\left(A^{1 / 3}\right)$
where $R_{0}$ is a constant whose value is found to be
$1.2 \times 10^{-15} m$ and $A$ is the mass number
$\therefore \frac{R_{X}}{R_{ Os }}=\left(\frac{A}{189}\right)^{1 / 3} $
$\Rightarrow \frac{1}{3}=\left(\frac{A}{189}\right)^{1 / 3} $
$A=\frac{189}{3^{3}}=\frac{189}{27}=7$
The given nucleus is $Li ^{7}$.