What is the product produced when $_{92}^{238} U$ $""_92^238"U"$ undergoes alpha decay?

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What is the product produced when $_{92}^{238} U$ $""_92^238"U"$ undergoes alpha decay?

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Answer 1 · 2018-05-17T00:23:47

Thorium- $234$ $234$ and an alpha particle of course!

Explanation:

The equation for alpha decay is:

$_{Z}^{A} X \to_{Z - 2}^{A - 4} X$ $""_(Z)^(A)X -> ""_(Z-2)^(A-4)X$ + $_{2}^{4} α$ $""_2^4 alpha$ ,

where:

$Z$ $"Z"$ is the atomic number, $A$ $"A"$ is the mass number, $X$ $"X"$ is the symbol of the element, and $α$ $alpha$ is a helium-4 nucleus.

So here, the new mass number of the atom will be $238 - 4 = 234$ $238-4=234$ , and the new element will have atomic number of $92 - 2 = 90$ $92-2=90$ , which is thorium.

So, the equation for the alpha decay of uranium is:

$_{92}^{238} U \to_{90}^{234} T h +_{2}^{4} α$ $""_92^238U->""_90^234Th+""_2^4alpha$

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What is the product produced when $_{92}^{238} U$ $""_92^238"U"$ undergoes alpha decay?

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What is the product produced when 23892U""_92^238"U" undergoes alpha decay?

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What is the product produced when $_{92}^{238} U$ $""_92^238"U"$ undergoes alpha decay?