1.

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

Answer» Our first step will be to write the reaction, then find the disintegration energy `Q`. If Q lr 0, the decay is allowed.
`alpha-` decay: `(._(89)^(226)) Ac rArr .(_(87)^(222)Fr) + alpha`
`Q=[M(._89)^(226)Ac)-M (._(87)^(222)Fr)-M.^(4He)]c^(2)`
`=5.50 MeV`
`beta-` deacy: `._(89)^(226) Ac) rarr (._(90)^(226)Th )+beta- + bar(V)`
`Q=[M(._89)^(226)Ac)-M (._(90)^(226)Th)]c^(2)`
`1.12 MeV`
`beta+` decay: `(._(89)^(226)AC)rarr._(88)^(226)Ra) +beta+ +v`
`Q=[M(._89)^(226)Ac)-M (._(88)^(226)Ra)-2m_(e))]c^(2)`
`=-0.38 MeV`
Electron capture : `(._(89)^(226)Ac) + e rarr ._(88)^(226)Ra +v`
`Q=[M(._89)^(226)Ac)-M (._(88)^(226)Ra)]c^(2)`
`=0.64 MeV` (Electron capture is allowed).


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