
Jem P. answered 11/06/15
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In Beta minus (β−) decay, mass number (top) A remains the same, atomic number (bottom) Z increases by one, and an electron is emitted and electron antineutrino is emitted.
42
19K →
42
20Ca + e− + ν-e(electron antineutrino= equal mass but opposite sign)
19K →
42
20Ca + e− + ν-e(electron antineutrino= equal mass but opposite sign)
52
26Fe →
52
27Co + e− + ν-e
26Fe →
52
27Co + e− + ν-e
14
6C →
14
7N + e− + ν-e
7N + e− + ν-e
16
7N →
16
8O + e− + ν-e
7N →
16
8O + e− + ν-e
Gamma ray emission is accompanied by β− decay in an excited state (*), which then drops down to a ground state of the atom (lowest energy state)
234
90Th →
234
91Pa* + e− + ν-e + γ +Energy release (specific to atom)
234
91Pa*→
234
91Pa + γ +Energy release (specific to atom)
90Th →
234
91Pa* + e− + ν-e + γ +Energy release (specific to atom)
234
91Pa*→
234
91Pa + γ +Energy release (specific to atom)
In Beta plus (β+) decay, mass number (top) A remains the same, atomic number (bottom) Z decreases by one, and a positron (e+) is emitted and electron neutrino is emitted.
42
19K →
42
18Ar + e+ + νe(electron neutrino= equal mass but opposite sign)
52
26Fe →
52
25Mn + e+ + νe
14
6C →
14
5B + e+ + νe
16
7N →
16
6C + e+ + νe
Technically gamma ray problem does not apply here, because usually accompanied with β- and not β+ decay.
42
19K →
42
18Ar + e+ + νe(electron neutrino= equal mass but opposite sign)
52
26Fe →
52
25Mn + e+ + νe
14
6C →
14
5B + e+ + νe
16
7N →
16
6C + e+ + νe
Technically gamma ray problem does not apply here, because usually accompanied with β- and not β+ decay.