Show that `(._(26)^(55)Fe)` may electron capture, but not `beta^(+)` decay.
Masses given are `M(._(26)^(55)Fe)=54.938298 am u`,
`M(._(25)^(55)Mn)=54.938050 am u, m(e )=0.000549 am u`.

Explanation

The two possible reactions for electron capture and `beta^(+)` decay by `(._(26)^(55)Fe)` are
Electron capture: `(._(26)^(55)Fe)`+ e rarr `(._(25)^(55)Mn)+n`
`beta^(+)` decay:`(._(26)^(55)Fe) rarr (._(25)^(55)Mn)` +beta^(+) +v`
First, we will determine the disintegration energy `Q` of `Eqs`. and (ii).
`M(._(26)^(55)Fe) =54.938298 a.m.u.`
`M(._(25)^(55)Mn)=54.938050 a.m.u.` and `m_(e)=0.000549 a.m.u`
For `beta^(+)` decay, the `Q` value of the reaction is `Q =[54.93.8298 -54.938050 - 2 xx 0.000549]`
`xx 931.5 MeV`
`= -0.79 MeV` Negative value of `Q` implies that positron decay is not possible spontaneously . For electron capture, the `Q` value of raction is
`Q=[54.938298 -54.938050] xx 931.5`
`=0.23 MeV`.