Nuclei - Result Question 4

4. 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: 4. (d)

Solution:

  1. (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$

$\frac{R_2}{R_1}=\frac{5 R_0}{3 R_0}=\frac{5}{3} R_1=\frac{5}{3} \times 3.6=6 fm$