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Nuclear size
Appears in
Concepts tested here
- R proportional to A^(1/3) 2
All Questions
2015 AIPMT-I 1 question
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If radius of the $^{27}_{13}Al$ nucleus is taken to be $R_{Al}$ then the radius of $^{125}_{53}Te$ nucleus is nearly:Nuclear radius $R \propto A^{1/3}$
$\frac{R_{Te}}{R_{Al}} = \left(\frac{125}{27}\right)^{1/3} = \frac{5}{3}$
$R_{Te} = \frac{5}{3}R_{Al}$
2012 AIPMT-PRE 1 question
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If the nuclear radius of $^{27}Al$ is 3.6 Fermi, the approximate nuclear radius of $^{64}Cu$ in Fermi is:$R = R_0A^{1/3}$, so $R \propto A^{1/3}$
$\frac{R_{Al}}{R_{Cu}} = \left(\frac{27}{64}\right)^{1/3} = \frac{3}{4}$
$\frac{3.6}{R_{Cu}} = \frac{3}{4} \Rightarrow R_{Cu} = 4.8$ Fermi