If radius of the $_{13} A l^{27}$ nucleus is estimated to be $3.6 f e r m i$ , then the radius of $_{52} T e^{125}$ nucleus nearly

6 f e r m i

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8 f e r m i

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4 f e r m i

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5 f e r m i

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Solution

The correct option is C $6 f e r m i$
We know that:
Radius of nucleus $\propto$ (Atomic mass number) $^{\frac{1}{3}}$
Radius $= K A^{\frac{1}{3}}$ where $K$ is constant of proportionality.
$\Rightarrow$ Radius $= K (27)^{\frac{1}{3}}$ [for $A l$ ]
$\Rightarrow 3.6 = 3 K$
$\Rightarrow K = 1.2$
For $T e$ , Radius $= (1.2) (125)^{\frac{1}{3}} = 1.2 \times 5 = 6 f e r m i$

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_{13}^{27} A l

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_{52}^{125} T e

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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:

_{13}^{27} A l

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_{52}^{125} T e

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