Question

If radius of the ${}_{13}^{27}Al$ nucleus is taken to be $R_{Al}$, then the radius of ${}_{52}^{125}Te$ nucleus is nearly:

• A. $\frac{3}{5}{R}_{Al}$
• B. ${\left(\frac{13}{53}\right)}^{\frac{1}{3}}{R}_{Al}$
• C. ${\left(\frac{53}{13}\right)}^{\frac{1}{3}}{R}_{Al}$
• D. $\frac{5}{3}{R}_{Al}$

Answer 1

Answer: (D)

$\frac{5}{3}{R}_{Al}$

$R=R_0{A}^{\frac{1}{3}}$
$\therefore \frac{R_{Al}}{R_{Te}}=(\frac{27}{125}{)}^{\frac{1}{3}}=\frac{3}{5}$
$R_{Te}=\frac{5}{3}{R}_{Al}$