In the successive emission of alpha- and beta | vedclass

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Question

(A) The nuclear reaction is: ${}^{238}_{92}U \longrightarrow {}^{206}_{82}Pb + m({}^{4}_{2}He) + n({}^{0}_{-1}e)$.Comparing the mass numbers on both s

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$8 \alpha, 6 \beta$

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(A) The nuclear reaction is: ${}^{238}_{92}U \longrightarrow {}^{206}_{82}Pb + m({}^{4}_{2}He) + n({}^{0}_{-1}e)$.Comparing the mass numbers on both sides: $238 = 206 + 4m$ $\Rightarrow 4m = 32$ $\Rightarrow m = 8$.Comparing the atomic numbers on both sides: $92 = 82 + 2m - n$.Substituting $m = 8$: $92 = 82 + 2(8) - n$ $\Rightarrow 92 = 82 + 16 - n$ $\Rightarrow 92 = 98 - n$ $\Rightarrow n = 6$.Therefore,$8$ $\alpha$-particles and $6$ $\beta$-particles are emitted.

In the successive emission of $\alpha$- and $\beta$-particles,the number of $\alpha$- and $\beta$-particles that should be emitted for the conversion of ${}^{238}_{92}U$ to ${}^{206}_{82}Pb$ are:

Similar Questions

In the following nuclear reaction,the other product is
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In the radioactive disintegration series ${}_{90}^{232}Th \rightarrow {}_{82}^{208}Pb$ involving $\alpha$ and $\beta$ decay,the total number of $\alpha$ and $\beta$ particles emitted are:

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