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13.Nuclei
medium
If radius of the $^{27}_{13}Al$ nucleus is taken to be $R_{Al}$, then the radius of $^{125}_{53}Te$ nucleus is nearly
A
${\left( {\frac{{53}}{{13}}} \right)^{\frac{1}{3}}}{R_{Al}}$
B
$\frac{5}{3}{R_{Al}}$
C
$\;\frac{3}{5}{R_{Al}}$
D
$\;{\left( {\frac{{13}}{{53}}} \right)^{\frac{1}{3}}}{R_{Al}}$
(AIPMT-2015) (AIPMT-1990)
Solution
Radius of the nucleus $R\,=\,R_{0} A^{1 / 3}$
$\therefore$ $\frac{R_{\mathrm{Al}}}{R_{\mathrm{Te}}}=\left(\frac{A_{\mathrm{Al}}}{A_{\mathrm{Te}}}\right)^{1 / 3}$
Here , $A_{\mathrm{Al}} =27, A_{\mathrm{Te}}=125, R_{\mathrm{Te}}=? $
$\frac{R_{\mathrm{Al}}}{R_{\mathrm{Te}}}=\left(\frac{27}{125}\right)^{1 / 3}=\frac{3}{5} \Rightarrow \quad R_{\mathrm{Te}}=\frac{5}{3} R_{\mathrm{Al}}$
Standard 12
Physics
