Three α$$-particles$$ and one β$$-particle$$ decaying takes place in series from an isotope Ra $$236$$ $$88$$ . Finally the isotope obtained will be

By using nα$$=A-A$$'$$4$$ and nβ$$=2n$$α$$-Z+Z$$' ⇒A'$$=A-4n$$α$$=236-4$$ x $$3=224$$ and Z'$$=(n$$β$$-2n$$α$$+Z)=(1-2$$ x $$3+88)=83$$

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Question

Three $α$ -particles and one $β$ -particle decaying takes place in series from an isotope ${}_{88}{Ra}^{236}$ . Finally the isotope obtained will be

Solution

By using $n_{α} = \frac{A - A'}{4} and n_{β} = 2 n_{α} - Z + Z'$
$\Rightarrow A' = A - 4 n_{α} = 236 - 4 x 3 = 224 and Z' = (n_{β} - 2 n_{α} + Z) = (1 - 2 x 3 + 88) = 83$

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