Chemistry, 08.09.2020 14:01 awesomegrill

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.

Answers: 1

Show answers

Answers

Answer from: maddy6882

A) $^{191}_{79}Au + e^{-} \rightarrow ^{191}_{78}Pt + \nu_{e}$

B) $^{201}_{79}Au \rightarrow ^{201}_{80}Hg + e^{-} + \overline\nu_{e}}$

C) $^{198}_{79}Au \rightarrow ^{198}_{80}Hg + e^{-} + \overline\nu_{e}}$

D) $^{188}_{79}Au \rightarrow ^{188}_{78}Pt + e^{+} + \nu_{e}$

Explanation:

A) The reaction of electron capture is:

$^{A}_{Z-1}X + e^{-} \rightarrow ^{A}_{Z}Y + \nu_{e}$

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 ( $\nu_{e}$ ) is emitted.

Hence, for the Gold-191 we have:

$^{191}_{79}Au + e^{-} \rightarrow ^{191}_{78}Pt + \nu_{e}$

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

$^{201}_{79}Au \rightarrow ^{201}_{80}Hg + e^{-} + \overline\nu_{e}}$

The above reaction is a beta decay reaction, in which a positron and electron antineutrino ( $\overline\nu_{e}$ ) 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:

$^{198}_{79}Au \rightarrow ^{198}_{80}Hg + e^{-} + \overline\nu_{e}}$

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

D) Gold-188 decays by positron emission:

$^{188}_{79}Au \rightarrow ^{188}_{78}Pt + e^{+} + \nu_{e}$

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 ( $\nu_{e}$ ) are released.

I hope it helps you!

Answer from: Quest

hi my dude or dudette

Answer from: Quest

it can be observed without changing the identity of the object.

it depends on what substance you're talking about as well as many other factors. in quantum mechanics, the observation of particles affects the observers ability to measure certain properties. for example: you can know the velocity of a particle, but you can't know it's position; or you can know it's position, but not it's velocity. it's like trying to find out where a grain of sand is by throwing other grains of sand at it. once you find out where it is, it's been moved. it's also like trying to find an answer to the question of whether or not a tree falling in a forest makes a sound if nobody is there to hear it. paradoxically, the only way to find out is to be there and hear it.