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(B) The initial nucleus is ${ }_{83}^{214} Bi$.Emission of a $\beta$-particle $({ }_{-1}^{0} e)$ increases the atomic number by $1$ while the mass num

Vedclass · Accepted Answer

$128$

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(B) The initial nucleus is ${ }_{83}^{214} Bi$.Emission of a $\beta$-particle $({ }_{-1}^{0} e)$ increases the atomic number by $1$ while the mass number remains unchanged:${ }_{83}^{214} Bi \rightarrow { }_{84}^{214} Po + { }_{-1}^{0} e$Emission of an $\alpha$-particle $({ }_{2}^{4} He)$ decreases the mass number by $4$ and the atomic number by $2$:${ }_{84}^{214} Po \rightarrow { }_{82}^{210} Pb + { }_{2}^{4} He$The final nucleus is ${ }_{82}^{210} Pb$.The number of neutrons $N = A - Z = 210 - 82 = 128$.

After the emission of a $\beta$-particle followed by an $\alpha$-particle from ${ }_{83}^{214} Bi$, the number of neutrons in the resulting atom is:

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