Consider the radioactive disintegration {}_{8 | vedclass

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(D) The initial nucleus is ${}_{82}A^{210}$ and the final nucleus is ${}_{82}D^{206}$.The change in mass number is $210 - 206 = 4$ units,and the chang

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$\beta, \beta, \alpha$

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(D) The initial nucleus is ${}_{82}A^{210}$ and the final nucleus is ${}_{82}D^{206}$.The change in mass number is $210 - 206 = 4$ units,and the change in atomic number is $82 - 82 = 0$.Emission of one $\alpha$-particle decreases the mass number by $4$ and the atomic number by $2$.Emission of one $\beta$-particle keeps the mass number constant and increases the atomic number by $1$.To keep the atomic number constant while decreasing the mass number by $4$,we need one $\alpha$-particle ($-2$ in atomic number) and two $\beta$-particles ($+2$ in atomic number).The sequence is ${}_{82}A^{210}$ $\xrightarrow{-\beta} {}_{83}B^{210}$ $\xrightarrow{-\beta} {}_{84}C^{210}$ $\xrightarrow{-\alpha} {}_{82}D^{206}$.Thus,the sequence of emission is $\beta, \beta, \alpha$.

Consider the radioactive disintegration ${}_{82}A^{210}$ $\rightarrow B$ $\rightarrow C$ $\rightarrow {}_{82}D^{206}$. The sequence of emission can be

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