Question

Nuclear radius of ${}^{27}Al$ is $3.6\times {10}^{-15}\text{m}$, approximate radius of ${}^{64}Cu$ is

• A. $4.8\times {10}^{-15}\text{m}$
• B. $2.4\times {10}^{-15}\text{m}$
• C. $6.4\times {10}^{-15}\text{m}$
• D. $1.2\times {10}^{-15}\text{m}$

Answer 1

The correct option is A $4.8\times {10}^{-15}\text{m}$

We know that
$\frac{R_1}{R_2}={\left(\frac{A_1}{A_2}\right)}^{\frac{1}{3}}\frac{R_1}{R_2}={\left(\frac{27}{64}\right)}^{\frac{1}{3}}\frac{R_1}{R_2}=\frac{3}{4}$
Given that $R_1=3.6\times {10}^{-15}\text{m}$
$R_2=\frac{4}{3}{R}_1=\frac{4}{3}\times 3.6\times {10}^{-15}=4.8\times {10}^{-15}\text{m}$