Find whether `alpha-` deacy or any of the `beta-` decay are allowed for `._(89)^(226)Ac`.

Answer

chance of `alpha`-decay are maximun

Explanation

Our first step will be to write the reaction, then to find the disintegration energy `Q`. if `Q le 0`, the decay is allowed. ltbgt Alpha decay: `(._(89)^(226)Ac) rarr (._(87)^(222)Fr ) + alpha`
`Q =[M(._(89)^(226)Ac) -M(._(87)^(222)Fr) -M (.^(4)He)]c^(2)`
`=5.50 MeV" "(beta^(-) decay is allowed)`
`beta` decay: `._(89)^(226)Ac=._(90)^(226)Th +veca+vecv`
`Q=[M(._(89)^(226)Ac)-M(._(90)^(226)Th)]c^(2)`
`=1.12 MeV" "(beta^(-)"decay is allowed")`
`beta^(bar)` decay: `._(89)^(226)Ac rarr ._(88)^(226)Ra +beta^(+) +v`
`Q=[M(._(89)^(226)Ac) -M(._(89)^(226) A) (._(88)^(226)Ac)Ra -2 m_(e)]c^(2)`
` = -0.38 MeV" "(beta^(+)" decay is not allowed")`
Electron capture: `._(89)^(226)Ac) +bar(e^(-)) rarr ._(88)^(226)Ra) +v`
`Q=[ M(._(89)^(226)Ac) -M(._(88)^(226)Ra)]c^(2)`
`0.64 MeV" "("Electron capture is allowed")`
In the above analysis, it is clear that during `alpha` decay, `Q-`value is maximum. Thus, chance of `alpha`-decay are maximum.