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

Option: 1

\left ( \frac{53}{13} \right )^{\frac{1}{3}} R_{Al}


Option: 2

\left ( \frac{5}{3} \right ) R_{Al}


Option: 3

\left ( \frac{3}{5} \right ) R_{Al}


Option: 4

\left ( \frac{13}{3} \right ) R_{Al}


Answers (1)

best_answer

 Radius of nucleus is given by

R=R_{0} A^{\frac{1}{3}}     

R\propto A^{\frac{1}{3}}

\frac{R_{Al}}{R_{Te}}=\left ( \frac{A_{Al}}{A_{Te}} \right )^{\frac{1}{3}}=\frac{3}{5}

R_{Te}=\frac{5}{3}R_{AI}

 

Posted by

jitender.kumar

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