The number of alpha and beta-particles emitte | vedclass

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Question

(A) For the nuclear reaction $_{90}^{228}Th 	o _{83}^{212}Bi$:$1$. The number of $\alpha$-particles emitted is calculated by the change in mass numbe

Vedclass · Accepted Answer

$4, 1$

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(A) For the nuclear reaction $_{90}^{228}Th 	o _{83}^{212}Bi$:$1$. The number of $\alpha$-particles emitted is calculated by the change in mass number:$	ext{Number of } \alpha = \frac{228 - 212}{4} = \frac{16}{4} = 4$.$2$. The number of $\beta$-particles emitted is calculated by the change in atomic number,considering the effect of $\alpha$-particles:$	ext{Number of } \beta = (	ext{Initial Atomic Number} - 	ext{Final Atomic Number}) - 2 	imes (	ext{Number of } \alpha)$$= (90 - 83) - 2 	imes 4 = 7 - 8 = -1$.Wait,let us re-evaluate the atomic number change: $90 - 2 	imes 4 + \beta = 83 \implies 90 - 8 + \beta = 83 \implies 82 + \beta = 83 \implies \beta = 1$.Therefore,$4$ $\alpha$-particles and $1$ $\beta$-particle are emitted.

The number of $\alpha$ and $\beta$-particles emitted in the nuclear reaction $_{90}^{228}Th \to _{83}^{212}Bi$ are respectively:

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