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(A) The isotope $_{83}^{214}Bi$ is a member of the Uranium series (also known as the Radium series). In the radioactive decay chain of $_{83}^{214}Bi$

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$1$

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(A) The isotope $_{83}^{214}Bi$ is a member of the Uranium series (also known as the Radium series). In the radioactive decay chain of $_{83}^{214}Bi$,it undergoes $\beta$-decay to become $_{84}^{214}Po$. Subsequently,$_{84}^{214}Po$ undergoes $\alpha$-decay to become $_{82}^{210}Pb$. However,the question asks for the number of $\alpha$-particles emitted specifically from the decay of $_{83}^{214}Bi$ in its immediate transition. In the specific decay process of $_{83}^{214}Bi ightarrow _{84}^{214}Po + e^- + \bar{
u}_e$,no $\alpha$-particle is emitted. If the question refers to the total number of $\alpha$-particles emitted by the daughter product $_{84}^{214}Po$ in the sequence,it is $1$. Given the standard context of such physics problems regarding the decay chain,the answer is $1$.

How many energy rays of $\alpha$-particles are emitted from $_{83}^{214}Bi$?

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