By emitting an alpha particle, an atom of uranium-235 decays into an atom of thorium. An alpha particle, which is the same as a helium nucleus, is represented by $_{2}^{4} H e$ $"_2^4He$ . How would you complete the equation for this reaction? $_{92}^{235} U$ $"_92^235U$ $\to_{?}^{?} H e$ $-> "_?^?He$

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By emitting an alpha particle, an atom of uranium-235 decays into an atom of thorium. An alpha particle, which is the same as a helium nucleus, is represented by $_{2}^{4} H e$ $"_2^4He$ . How would you complete the equation for this reaction? $_{92}^{235} U$ $"_92^235U$ $\to_{?}^{?} H e$ $-> "_?^?He$

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Answer 1 · 2017-01-01T19:06:00

$_{92}^{235} U \to_{2}^{4} H e +_{90}^{231} T h$ $""_92^235U rarr ""_2^4He + ""_90^231Th$

Explanation:

I can't say that mass and charge are conserved here, because this is a nuclear reaction not a chemical one. I don't know how energy is transferred here; nor do I know the half-life of $_{92}^{235} U$ $""_92^235U$ . You might get a better answer in the Physics section.......

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By emitting an alpha particle, an atom of uranium-235 decays into an atom of thorium. An alpha particle, which is the same as a helium nucleus, is represented by 42He"_2^4He. How would you complete the equation for this reaction? 23592U"_92^235U→??He -> "_?^?He

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