The number of alpha and beta-particles emitte | vedclass

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Question

(A) The nuclear reaction is $_{92}U^{238} 	o _{90}Th^{234} 	o _{91}Pa^{234}$.Step $1$: For the emission of $\alpha$-particles,the mass number decrea

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$1$ and $1$

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(A) The nuclear reaction is $_{92}U^{238} 	o _{90}Th^{234} 	o _{91}Pa^{234}$.Step $1$: For the emission of $\alpha$-particles,the mass number decreases by $4$ and the atomic number decreases by $2$. Change in mass number $= 238 - 234 = 4$. Number of $\alpha$-particles $= \frac{4}{4} = 1$.Step $2$: For the emission of $\beta$-particles,the atomic number increases by $1$ while the mass number remains constant.After the emission of $1$ $\alpha$-particle,the atomic number becomes $90$. To reach $91$ $(Pa)$,the atomic number must increase by $1$,which corresponds to the emission of $1$ $\beta$-particle.Therefore,the number of $\alpha$ and $\beta$-particles emitted are $1$ and $1$ respectively.

The number of $\alpha$ and $\beta$-particles emitted in the nuclear reaction $_{92}U^{238} \to _{90}Th^{234} \to _{91}Pa^{234}$ are respectively:

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