If radius of the ^{27}_{13}Al nucleus is | vedclass

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Question

Radius of the nucleus  $R\,=\,R_{0} A^{1 / 3}$

$	herefore$  $\frac{R_{\mathrm{Al}}}{R_{\mathrm{Te}}}=\left(\frac{A_{\mathrm{Al}}}{A_{\ma

Vedclass · Accepted Answer

$\frac{5}{3}{R_{Al}}$

Vedclass · Answer

Radius of the nucleus  $R\,=\,R_{0} A^{1 / 3}$

$	herefore$  $\frac{R_{\mathrm{Al}}}{R_{\mathrm{Te}}}=\left(\frac{A_{\mathrm{Al}}}{A_{\mathrm{Te}}}ight)^{1 / 3}$

Here , $A_{\mathrm{Al}} =27, A_{\mathrm{Te}}=125, R_{\mathrm{Te}}=? $

$\frac{R_{\mathrm{Al}}}{R_{\mathrm{Te}}}=\left(\frac{27}{125}ight)^{1 / 3}=\frac{3}{5} \Rightarrow \quad R_{\mathrm{Te}}=\frac{5}{3} R_{\mathrm{Al}}$

If radius of the $^{27}_{13}Al$ nucleus is taken to be $R_{Al}$, then the radius of $^{125}_{53}Te$ nucleus is nearly

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