Question

If radius of the ${}_{13}A{l}^{27}$ nucleus is estimated to be $3.6fermi$, then the radius of ${}_{52}T{e}^{125}$ nucleus nearly

• A. $6fermi$
• B. $8fermi$
• C. $4fermi$
• D. $5fermi$

Answer 1

Answer: (A)

$6fermi$

We know that:
Radius of nucleus $\propto$ (Atomic mass number)${}^{\frac{1}{3}}$

Radius $=K{A}^{\frac{1}{3}}$ where $K$ is constant of proportionality.
$\Rightarrow$ Radius $=K(27{)}^{\frac{1}{3}}$ [for $Al$]
$\Rightarrow 3.6=3K$

$\Rightarrow K=1.2$

For $Te$, Radius $=\left(1.2\right)(125{)}^{\frac{1}{3}}=1.2\times 5=6fermi$