Answer:
A) [tex]^{191}_{79}Au + e^{-} \rightarrow ^{191}_{78}Pt + \nu_{e}[/tex]
B) [tex]^{201}_{79}Au \rightarrow ^{201}_{80}Hg + e^{-} + \overline\nu_{e}}[/tex]
C) [tex]^{198}_{79}Au \rightarrow ^{198}_{80}Hg + e^{-} + \overline\nu_{e}}[/tex]
D) [tex] ^{188}_{79}Au \rightarrow ^{188}_{78}Pt + e^{+} + \nu_{e} [/tex]
Explanation:
A) The reaction of electron capture is:
[tex] ^{A}_{Z-1}X + e^{-} \rightarrow ^{A}_{Z}Y + \nu_{e} [/tex]
Where:
A: is the mass number = n + Z
Z: is the number of protons
n: is the number of neutrons
In electron capture reaction a proton of the atom is converted into a neutron and an electron neutrino ([tex]\nu_{e}[/tex]) is emitted.
Hence, for the Gold-191 we have:
[tex]^{191}_{79}Au + e^{-} \rightarrow ^{191}_{78}Pt + \nu_{e}[/tex]
B) The nuclear reaction of the decay of Au-201 to Hg-201 is:
[tex]^{201}_{79}Au \rightarrow ^{201}_{80}Hg + e^{-} + \overline\nu_{e}}[/tex]
The above reaction is a beta decay reaction, in which a positron and electron antineutrino ([tex]\overline\nu_{e}[/tex]) are emitted from the Au-201 nucleus and a neutron is converted to a proton (n-1 and Z+1; A remains constant).
C) The nuclear reaction of beta emission Au-198 is:
[tex]^{198}_{79}Au \rightarrow ^{198}_{80}Hg + e^{-} + \overline\nu_{e}}[/tex]
Same as above, a beta emission produces the emission of a positron and electron antineutrino ([tex]\overline\nu_{e}[/tex]) and the conversion of a neutron into a proton.
D) Gold-188 decays by positron emission:
[tex] ^{188}_{79}Au \rightarrow ^{188}_{78}Pt + e^{+} + \nu_{e} [/tex]
In a positron emission, we have that a proton of Au-188 is converted into a neutron and a positron (e⁺) and an electron neutrino ([tex]\nu_{e}[/tex]) are released.
I hope it helps you!