An element E loses one alpha and two beta par | vedclass

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(C) Let the initial element be ${}_{Z}E^{M}$, where $Z$ is the atomic number and $M$ is the atomic mass.When an $\alpha$-particle $({}_{2}He^{4})$ is

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an isotope of $E$

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(C) Let the initial element be ${}_{Z}E^{M}$, where $Z$ is the atomic number and $M$ is the atomic mass.When an $\alpha$-particle $({}_{2}He^{4})$ is emitted, the atomic number decreases by $2$ and the atomic mass decreases by $4$: ${}_{Z}E^{M} \rightarrow {}_{Z-2}E'^{M-4} + {}_{2}He^{4}$.When two $\beta$-particles $({}_{-1}e^{0})$ are emitted, the atomic number increases by $1$ for each $\beta$-particle (total increase of $2$), while the atomic mass remains unchanged: ${}_{Z-2}E'^{M-4} \rightarrow {}_{Z}E''^{M-4} + 2({}_{-1}e^{0})$.The final element is ${}_{Z}E''^{M-4}$.Since the final element has the same atomic number $Z$ as the initial element $E$ but a different atomic mass $M-4$, it is an isotope of $E$.

An element $E$ loses one $\alpha$ and two $\beta$ particles in successive stages. The resulting element will be

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