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

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

Text Solution

Verified by Experts

`alpha`-decay : `""_zX^(A) to ""_(z-2)Y^(A-4) + ""_2He^(4)`
`3alpha`-decay:
`""_zX^(A) to ""_(z-6)Y^(A-12) + 3(""_2He^(4))`
`therefore ""_88Ra^(226) to ""_82Pb^(214) +3""_2He^(4) (or)`
`""_88Ra^(226) to ""_82X^(214) +3""_2He^(4)`

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During a - decay the mass number of the daughter element is:

more by 2 units

less by 2 units

more by 4 units

less by 4 units

In _88Ra^(226) nucleus , there are :

138 protons and 88 neutrons

138 neutrons and 88 protons

226 protons and 88 electrons

226 neutrons and 138 electrons

When a - particle is emitted the atomic number of daughter element is: