If the nucleus {}_{13}^{27}Al has a nucl | vedclass

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Question

Nuclear radii $R=\left(R_{0}ight) A^{1 / 3}$

where $A$ is the mass number.

$	herefore \frac{{{R_{{	ext{Ge}}}}}}{{{R_{{	ext{Al}}}}}} = {\lef

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Nuclear radii $R=\left(R_{0}ight) A^{1 / 3}$

where $A$ is the mass number.

$	herefore \frac{{{R_{{	ext{Ge}}}}}}{{{R_{{	ext{Al}}}}}} = {\left( {\frac{{{A_{{	ext{Ge}}}}}}{{{A_{{	ext{Al}}}}}}} ight)^{1/3}}$ $ = {\left( {\frac{{125}}{{27}}} ight)^{1/3}} = \left( {\frac{5}{3}} ight)$

or, $R_{Ge}=\frac{5}{3} 	imes R_{A l}=\frac{5}{3} 	imes 3.6=6 \mathrm{\,fm}$

$(Given \;R_{\mathrm{Al}}=3.6 \mathrm{\,fm})$

If the nucleus ${}_{13}^{27}Al$ has a nuclear radius of about $3.6\,\, fm,$ then ${ }_{32}^{125} Te$ would have its radius approximately as .......$fm$

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