In the radioactive decay series, ${}^{238}_{92}U$ ultimately decays to ${}^{206}_{82}Pb$. How many $\alpha$ and $\beta^-$ particles are emitted?
In the radioactive decay series, ${}^{238}_{92}U$ ultimately decays to ${}^{206}_{82}Pb$. How many $\alpha$ and $\beta^-$ particles are emitted?
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 2mo ago
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Let $\alpha$ particles emitted $= x$, $\beta^-$ particles emitted $= y$.
From mass number: $238 - 4x = 206 \Rightarrow x = 8$
From atomic number: $92 - 2x + y = 82 \Rightarrow 92 - 16 + y = 82 \Rightarrow y = 6$
$$8 alpha particles and 6 beta particles are emitted$$
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Discussion (3)
KS
I solved it a slightly different way and got the same answer, good sign.
N
Thanks a ton, I was stuck on this exact problem for an hour.
NA
Why do we take the positive value only in the last step?