Q157 / 160 · TS EAMCET 2013 · Physics
The radius of ${ }_{72} \mathrm{Te}^{125}$ nucleus is 6 fermi. The radius of ${ }_{13} \mathrm{Al}^{27}$ nucleus in meters is
- A$3.6 \times 10^{-12} \mathrm{~m}$
- Correct answerB$3.6 \times 10^{-15} \mathrm{~m}$
- C$7.2 \times 10^{-8} \mathrm{~m}$
- D$7.2 \times 10^{-15} \mathrm{~m}$
Solution
The relation between radius $(R)$ and atomic number $(A)$ is
$\frac{R_1}{R_2}=\left(\frac{A_1}{A_2}\right)^{1 / 3}$
Given, $R_1=6$ fermi, $A_1=125, A_2=27$
$\begin{aligned}
\frac{6}{R_2} & =\left(\frac{125}{27}\right)^{1 / 3}=\frac{5}{3} \\
\Rightarrow \quad \quad \quad R_2 & =\frac{6 \times 3}{5}=\frac{18}{5}=3.6 \text { fermi } \\
& =3.6 \times 10^{-15} \mathrm{~m}
\end{aligned}$
Hence, the correct answer is option: 'B'.