Nuclei - Result Question 1
1. If radius of the $ _12^{27} Al$ nucleus is taken to be $R _{Al}$, then the radius of $ _53^{125} Te$ nucleus is nearly:
[2015]
(a) $\frac{5}{3} R _{Al}$
(b) $\frac{3}{5} R _{Al}$
(c) $(\frac{13}{53})^{1 / 3} R _{Al}$
(d) $(\frac{53}{13})^{1 / 3} R _{Al}$
Show Answer
Solution:
- (a) As we know, $R=R_0(A)^{1 / 3}$
where $A=$ mass number
$R _{AI}=R_0(27)^{1 / 3}=3 R_0$
$R _{Te}=R_0(125)^{1 / 3}=5 R_0=\frac{5}{3} R _{AI}$
