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

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Answer 1 · 2018-05-17T00:23:47

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

The equation for alpha decay is:

$""_(Z)^(A)X -> ""_(Z-2)^(A-4)X$ + $""_2^4 alpha$ ,

where:

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

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

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

$""_92^238U->""_90^234Th+""_2^4alpha$

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