A nucleus with atomic mass number ' $A$ ' produces another nucleus by loosing 2 alpha particles. The volume of the new nucleus is 60 times that of the alpha particle. The atomic mass number A of the original nucleus is

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Aicharya · Accepted Answer

${ }_Z^A X \xrightarrow{-\infty-	ext { particle }}{ }_{\mathrm{Z}-2}^{\mathrm{A}-4} \mathrm{Y} \xrightarrow{-\infty-	ext { particle }}{ }_{\mathrm{Z}-4}^{\mathrm{A}-8} \mathrm{R}$ Volume of $\mathrm{R}=60\left(ight.$ Volume of $\left.{ }_2^4 \mathrm{He}ight)$ $\Rightarrow \frac{4}{3} \pi r_1^3=60 	imes \frac{4}{3} \pi r_2^3 \Rightarrow\left(R_0 A_1^{\frac{1}{3}}ight)^3=60\left(R_0 A_2^{\frac{1}{3}}ight)^3$ $\Rightarrow(\mathrm{A}-8)=60 	imes 4 \quad 	herefore \quad A=248$

A nucleus with atomic mass number ' $A$ ' produces another nucleus by loosing 2 alpha particles. The volume…

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