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(B) The initial atom is $_{90}X^{232}$.Step $1$: Emission of two $\beta$-particles.Each $\beta$-particle emission increases the atomic number by $1$ b

Vedclass · Accepted Answer

$5$

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(B) The initial atom is $_{90}X^{232}$.Step $1$: Emission of two $\beta$-particles.Each $\beta$-particle emission increases the atomic number by $1$ but does not change the mass number.After $2\beta$-emission: $_{90}X^{232} 	o _{92}Y^{232} + 2_{-1}e^{0}$.Step $2$: Emission of $x$ $\alpha$-particles to reach mass number $212$ and atomic number $82$.Each $\alpha$-particle $(_{2}He^{4})$ decreases the mass number by $4$ and the atomic number by $2$.Change in mass number: $232 - 212 = 20$.Number of $\alpha$-particles $(x)$ $= \frac{20}{4} = 5$.Verification of atomic number: $92 - (5 	imes 2) = 92 - 10 = 82$. This matches the target atomic number.Therefore,the number of $\alpha$-particles emitted is $5$.

An atom has a mass number of $232$ and an atomic number of $90$. How many $\alpha$-particles should it emit after the emission of two $\beta$-particles,so that the resulting atom has a mass number of $212$ and an atomic number of $82$?

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