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(B) The emission of an alpha particle $(_{2}He^{4})$ results in a decrease of $2$ in the atomic number and $4$ in the mass number.For the reaction $_{

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$1$

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(B) The emission of an alpha particle $(_{2}He^{4})$ results in a decrease of $2$ in the atomic number and $4$ in the mass number.For the reaction $_{84}Po^{215} 	o _{82}Pb^{211} + x(_{2}He^{4})$:Change in atomic number = $84 - 82 = 2$.Change in mass number = $215 - 211 = 4$.Since one alpha particle causes a change of $2$ in atomic number and $4$ in mass number,the number of alpha particles emitted is $1$.

How many alpha particles are emitted in the nuclear transformation
$_{84}Po^{215} \xrightarrow{} _{82}Pb^{211}$?

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How many alpha particles are emitted in the nuclear transformation$_{84}Po^{215} \xrightarrow{} _{82}Pb^{211}$?