Question

Write nuclear reaction equation for
(i) `alpha` decay of `._88Ra^(226)` (ii) `alpha` decay of `._94Pu^(242)` (iii) `beta^(-)` decay of `._15P^(32)`
(iv) `beta^(-)` decay of `._83Bi^(210)` (v) `beta^(+)` decay of `._6C^(11)` (vi) `beta^(+)` decay of `._43Tc^(97)`
(vii) Electron capture of `._54Xe^(120)`.

Answer 1

`alpha` is a nucleus of helium `(""_(2)He^(4))` and `beta` is an electron `(e^(−) " for " beta^(−) and e^(+)" for " beta^(+))`. In every `alpha-decay` , there is a loss of 2 protons and 4 neutrons. In every `beta^(+)-decay`, there is a loss of 1 proton and a neutrino is emitted from the nucleus. In every `beta^(-)-decay`, there is a gain of 1 proton and an antineutrino is emitted from the nucleus.
For the given cases, the various nuclear reactions can be written as:
`(i) " "_(88)Ra^(226) to _(86)Rn^(222)+_(2)He^(4)`
`(ii) " "_(94)^(242)Pu to _(92)^(238)U+_(2)^(4)He`
`(iii) " "_(15)^(32)P to _(16)^(32)S+e^(-)+barv`
`(iv) " "_(83)^(210)B to _(83)^(210)PO+e^(-)+barv`
`(v) " "_(6)^(11)C to _(5)^(11)B+e^(+)+v`
`(vi) " "_(43)^(97)Tc to _(42)^(97)Mo +e^(+)+v`
`(vii) " "_(54)^(120)Xe+e^(+) to _(53)^(120)I + v`