Question

Part AGold-191 undergoes electron capture.Express your answer as a nuclear equation.Part BGold-201 decays to a mercury isotope.Express your answer as a nuclear equation.Part CGold-198 undergoes beta emission.Express your answer as a nuclear equation.Part DGold-188 decays by positron emission.Express your answer as a nuclear equation.

Answer 1

A)
`^(191)_(79)Au + e^(-) \rightarrow ^(191)_(78)Pt + \\u_(e)`

B)
`^(201)_(79)Au \rightarrow ^(201)_(80)Hg + e^(-) + \overline\\u_(e)}`

C)
`^(198)_(79)Au \rightarrow ^(198)_(80)Hg + e^(-) + \overline\\u_(e)}`

D)
`^(188)_(79)Au \rightarrow ^(188)_(78)Pt + e^(+) + \\u_(e)`

Step-by-step explanation:

A) The reaction of electron capture is:

`^(A)_(Z-1)X + e^(-) \rightarrow ^(A)_(Z)Y + \\u_(e)`

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 (
`\\u_(e)`) is emitted.

Hence, for the Gold-191 we have:

`^(191)_(79)Au + e^(-) \rightarrow ^(191)_(78)Pt + \\u_(e)`

B) The nuclear reaction of the decay of Au-201 to Hg-201 is:

`^(201)_(79)Au \rightarrow ^(201)_(80)Hg + e^(-) + \overline\\u_(e)}`

The above reaction is a beta decay reaction, in which a positron and electron antineutrino (
`\overline\\u_(e)`) 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:

`^(198)_(79)Au \rightarrow ^(198)_(80)Hg + e^(-) + \overline\\u_(e)}`

Same as above, a beta emission produces the emission of a positron and electron antineutrino (
`\overline\\u_(e)`) and the conversion of a neutron into a proton.

D) Gold-188 decays by positron emission:

`^(188)_(79)Au \rightarrow ^(188)_(78)Pt + e^(+) + \\u_(e)`

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 (
`\\u_(e)`) are released.

I hope it helps you!