First slide

Size of nucleus

Question

If radius of the ${}_{13}^{27}{Al}$ nucleus is taken to be $R_{Al}$ , then the radius of ${}_{53}^{125}{Te}$ nucleus is nearly

Easy

A

$\frac{3}{5} R_{Al}$
B

${(\frac{13}{53})}^{1 / 3} R_{Al}$
C

${(\frac{53}{13})}^{1 / 3} R_{Al}$
D

$\frac{5}{3} R_{Al}$

Solution

$Radius of the nucleus R = R_{0} A^{1 / 3}$
$∴ \frac{R_{Al}}{R_{Te}} = {(\frac{A_{Al}}{A_{Te}})}^{1 / 3}$
$Here, A_{Al} = 27, A_{Te} = 125, R_{Te} = ?$
$\frac{R_{Al}}{R_{Te}} = {(\frac{27}{125})}^{1 / 3} = \frac{3}{5} \Rightarrow R_{Te} = \frac{5}{3} R_{Al}$

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