Question

If radius of the ${}_{13}$${Al}^{27}$ nucleus is taken to be $R_A$. Then the radius of ${}_{53}$${Te}^{125}$ nucleus is nearly:

• A. (13/53)^1/3 R_A
• B. (53/13)^1/3 R_A
• C. (5/3) R_A
• D. (3/5) R_A

Answer 1

The correct option is C

(5/3) R_A

Radius of the nucleus goes as

$R\alpha A^{\frac{1}{3}}$ , where A is the atomic mass.

If $R_{Te}$ is the radius of the nucleus of telurium atom and $R_{A1}$is the radius of the nucleus of aluminium atom we have

$\frac{R_{Te}}{R_A}=\frac{(125{)}^{\frac{1}{3}}}{(27{)}^{\frac{1}{3}}}=\frac{5}{3}\therefore R_{Te}=\frac{5}{3}{R}_A$