If radius of the $^{27} A l_{13}$ nucleus is taken to be $R_{A l}$ , then the radius of $^{125} T_{53}$ nucleus is nearly.

${[\frac{13}{53}]}^{\frac{1}{3}} R_{A l}$

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${[\frac{53}{15}]}^{\frac{1}{3}} R_{A l}$

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$\frac{5}{3} R_{A l}$

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$\frac{3}{5} R_{A l}$

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Solution

The correct option is C
$\frac{5}{3} R_{A l}$

Radius of Nucleus $\propto A^{\frac{1}{3}}$
$∴ A = m a s s n u m b e r R_{A l} = (27)^{\frac{1}{3}} R_{T e} = (125)^{\frac{1}{3}} \frac{R_{A l}}{R_{T e}} = \frac{3}{5} R_{T e} = \frac{5}{3} R_{A l}$

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Similar questions

If radius of the $_{13}$ ${A l}^{27}$ nucleus is taken to be $R_{A}$ . Then the radius of $_{53}$ ${T e}^{125}$ nucleus is nearly:

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What will be the radius of $^{64} C u$ nucleus if the radius of $^{27} A l$ Al nucleus is $R_{A l}$ ?

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