To determine the identity of the parent atom that underwent radioactive decay, we need to understand the characteristics of the decay process.
Given the decay equation:
[tex]\[
\rightarrow \rightarrow \,^{129}_{54} \text{Xe} + \,^{0}_{-1} e
\][/tex]
we can observe that a beta particle (electron) is emitted. In beta decay:
- The atomic number of the parent atom increases by 1 because a neutron in the nucleus converts into a proton, emitting an electron (beta particle).
- The mass number of the parent atom remains unchanged because the total number of nucleons (protons and neutrons) does not change; only a neutron changes into a proton.
From the equation, we know:
- The resulting daughter atom is Xenon ([tex]\( ^{129}_{54} \text{Xe} \)[/tex]).
- The emitted particle is an electron ([tex]\( ^{0}_{-1} e \)[/tex]).
To find the parent atom:
1. Mass Number: The mass number remains unchanged, so the parent atom has the same mass number as the daughter atom, which is [tex]\( 129 \)[/tex].
2. Atomic Number: The atomic number of the parent atom before the decay must have been 1 less than the resulting daughter atom's atomic number. Since the daughter atom is Xenon with an atomic number of 54, the parent atom's atomic number must have been [tex]\( 54 + 1 = 55 \)[/tex].
Thus, the parent atom has:
- An atomic number of 55.
- A mass number of 129.
The element with an atomic number of 55 is Barium (Ba).
Therefore, the identity of the parent atom is [tex]\( ^{129}_{55} \text{Ba} \)[/tex].
