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

medium

Nuclei

2015

physics

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

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

Explanation

To solve this problem, we need to use the formula for the radius of a nucleus, which is given by: $\\ R = R_0 A^{1/3} \\$ where $R_0$

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$\left(\frac{13}{53}\right)^{1/3} R_{Al}$

$\left(\frac{53}{13}\right)^{1/3} R_{Al}$

is a constant and

A

is the mass number of the nucleus.

\\

Given:

\\

• For

^{27}_{13}Al,

the radius is

R_{Al} = R_0 \cdot 27^{1/3}. \\

• We need to find the radius of

^{125}_{53}Te. \\

The radius of

^{125}_{53}Te

is given by:

\\ R_{Te} = R_0 \cdot 125^{1/3} \\

To find the relation between

R_{Te}

and

R_{Al},

we take the ratio:

\\ \frac{R_{Te}}{R_{Al}} = \frac{R_0 \cdot 125^{1/3}}{R_0 \cdot 27^{1/3}} \\

Simplify the expression:

\\ \frac{R_{Te}}{R_{Al}} = \left(\frac{125}{27}\right)^{1/3} \\

Calculate

\left(\frac{125}{27}\right)^{1/3}: \\ 125 = 5^3

and

27 = 3^3 \\ \left(\frac{125}{27}\right)^{1/3} = \left(\frac{5^3}{3^3}\right)^{1/3} = \frac{5}{3} \\

Thus,

R_{Te} = \frac{5}{3} R_{Al}. \\

Therefore, the radius of

^{125}_{53}Te

is nearly

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

This corresponds to Option 1.