Nuclei Ques 37
37. If the nucleus $ _13^{27} Al$ has nuclear radius of about $3.6 fm$, then $ _32^{125} Te$ would have its radius approximately as
[2007]
(a) $9.6 fm$
(b) $12.0 fm$
(c) $4.8 fm$
(d) $6.0 fm$.
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Answer:
Correct Answer: 37.(d)
Solution:
- (d) It has been known that a nucleus of mass number $A$ has radius
$R=R_0 A^{1 / 3}$,
where $R_0=1.2 \times 10^{-15} m$
and $A=$ mass number
In case of $ _13^{27} \mathrm{A \ell}$, let nuclear radius be $R_1$
and for $ _32^{125} Te$, nuclear radius be $R_2$
For $ _13^{27} Al, R_1=R_0(27)^{1 / 3}=3 R_0$
For $ _32^{125} Te, R_2=R_0(125)^{1 / 3}=5 R_0$
$\frac{R_2}{R_1}=\frac{5 R_0}{3 R_0}=\frac{5}{3} R_1=\frac{5}{3} \times 3.6=6 fm$
BETA
