How do you complete the following alpha decay reaction?

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How do you complete the following alpha decay reaction?

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Answer 1 · 2016-07-02T08:59:34

$_{90}^{232} T h \to_{88}^{228} R a +_{2}^{4} H e$ $""_90^232 Th -> ""_88^228Ra + ""_2^4He$

Explanation:

The general notation of a nuclide ( $X$ $X$ ) is:

$_{Z}^{A} X$ $""_Z^AX$

In wich $Z$ $Z$ is the number of protons and $A$ $A$ the mass number (protons + neutrons).

In alpha decay the nucleus emits a particle that contains 2 protons and 2 neutrons, which is similar to the nucleus of helium. So a notation for the nuclide ( $Y$ $Y$ ) that is left after emitting an alpha particle ( $_{2}^{4} H e$ $""_2^4He$ ) is:

$_{Z - 2}^{A - 4} Y +_{2}^{4} H e$ $""_(Z-2)^(A-4)Y + ""_2^4He$

Now you can complete the equation given in the example:

$_{88 + 2}^{228 + 4} X =_{90}^{232} X$ $""_(88+2)^(228+4)X = ""_90^232X$

The last step is finding the nuclide that has 90 protons and 142 neutrons in a table of nuclides. This appears to be Thorium ( $T h$ $Th$ ).

This makes the equation complete:

$_{90}^{232} T h \to_{88}^{228} R a +_{2}^{4} H e$ $""_90^232 Th -> ""_88^228Ra + ""_2^4He$

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How do you complete the following alpha decay reaction?

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