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If the nucleus $^{27}_{13}Al$ has nuclear radius of about 3.6 fm, then $^{125}_{52}Te$ would have its radius approximately as
A
9.6 fm
B
12.0 fm
C
4.8 fm
D
6.0 fm
Detailed Solution
A nucleus of mass number $A$ has radius $R = R_0 A^{1/3}$, where $R_0 = 1.2 \times 10^{-15}$ m.
For $^{27}Al$, let the nuclear radius be $R_1$: $R_1 = R_0 (27)^{1/3} = 3R_0$
For $^{125}Te$, let the nuclear radius be $R_2$: $R_2 = R_0 (125)^{1/3} = 5R_0$
$\dfrac{R_2}{R_1} = \dfrac{5R_0}{3R_0} = \dfrac{5}{3}$
$R_2 = \dfrac{5}{3} \times R_1 = \dfrac{5}{3} \times 3.6$
$R_2 = 6$ fm
For $^{27}Al$, let the nuclear radius be $R_1$: $R_1 = R_0 (27)^{1/3} = 3R_0$
For $^{125}Te$, let the nuclear radius be $R_2$: $R_2 = R_0 (125)^{1/3} = 5R_0$
$\dfrac{R_2}{R_1} = \dfrac{5R_0}{3R_0} = \dfrac{5}{3}$
$R_2 = \dfrac{5}{3} \times R_1 = \dfrac{5}{3} \times 3.6$
$R_2 = 6$ fm