_{95}Am^{241} and _{90}Th^{234} belong respec | vedclass

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(B) The radioactive series are classified based on the remainder when the mass number $(A)$ is divided by $4$:$1$. For $_{95}Am^{241}$,the mass number

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$4n + 1$ and $4n + 2$ radioactive disintegration series

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(B) The radioactive series are classified based on the remainder when the mass number $(A)$ is divided by $4$:$1$. For $_{95}Am^{241}$,the mass number is $241$. Dividing $241$ by $4$ gives a remainder of $1$ $(241 = 4 	imes 60 + 1)$,so it belongs to the $4n + 1$ series (Neptunium series).$2$. For $_{90}Th^{234}$,the mass number is $234$. Dividing $234$ by $4$ gives a remainder of $2$ $(234 = 4 	imes 58 + 2)$,so it belongs to the $4n + 2$ series (Uranium series).Therefore,the correct option is $B$.

$_{95}Am^{241}$ and $_{90}Th^{234}$ belong respectively to

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