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""88Ra^(226) experiences three alpha-dec...

`""_88Ra^(226)` experiences three `alpha`-decay . Find the number of neutrons in the daughter element.

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`""_(88)Ra^(226)` consider as a parent element that is `""_(88)X^(226)` and their daughter clement is `""_(Z)Y^(A)`
According to a decay process,
`""_(88)X^(226) overset(3 alpha "decay") ""_(82)Y^(214)+ 3alpha "decay"`
During the 3a decay, the atomic number decreases by 6 and mass number decreases by 12.
So the number of neutrons in the daughter element N=A-Z
N = 214 -88 = 126
Number of neutrons in the daughter element N = 126
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