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

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

Text Solution

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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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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: