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13.Nuclei
medium
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 .......$fm$
A
$9.6 $
B
$12 $
C
$4.8 $
D
$6 $
(AIPMT-2007) (AIEEE-2005)
Solution
Nuclear radii $R=\left(R_{0}\right) A^{1 / 3}$
where $A$ is the mass number.
$\therefore \frac{{{R_{{\text{Ge}}}}}}{{{R_{{\text{Al}}}}}} = {\left( {\frac{{{A_{{\text{Ge}}}}}}{{{A_{{\text{Al}}}}}}} \right)^{1/3}}$ $ = {\left( {\frac{{125}}{{27}}} \right)^{1/3}} = \left( {\frac{5}{3}} \right)$
or, $R_{Ge}=\frac{5}{3} \times R_{A l}=\frac{5}{3} \times 3.6=6 \mathrm{\,fm}$
$(Given \;R_{\mathrm{Al}}=3.6 \mathrm{\,fm})$
Standard 12
Physics
