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:

  1. (a) As we know, $R=R_0(A)^{1 / 3}$

where $A=$ mass number

$R _{AI}=R_0(27)^{1 / 3}=3 R_0$

$R _{Te}=R_0(125)^{1 / 3}=5 R_0=\frac{5}{3} R _{AI}$