To determine the type of radioactive decay that occurs when polonium-210 ([tex]\(^{210}_{84}\text{Po}\)[/tex]) decays to lead-206 ([tex]\(^{206}_{82}\text{Pb}\)[/tex]) with the emission of a helium nucleus ([tex]\(^{4}_{2}\text{He}\)[/tex]), we must analyze the given nuclear reaction:
[tex]\[_{84}^{210} \text{Po} \rightarrow \, _{82}^{206} \text{Pb} + \, _{2}^{4} \text{He}\][/tex]
Here, we can observe the following changes:
1. Atomic Number Change:
- Polonium (Po) has an atomic number of 84.
- After the decay, Lead (Pb) has an atomic number of 82.
- The emitted particle, a helium nucleus (He), has an atomic number of 2.
The total atomic number before the decay is [tex]\( 84 \)[/tex].
The total atomic number after the decay is [tex]\( 82 + 2 = 84 \)[/tex], which confirms the conservation of atomic number.
2. Mass Number Change:
- Polonium (Po) has a mass number of 210.
- After the decay, Lead (Pb) has a mass number of 206.
- The emitted particle, a helium nucleus (He), has a mass number of 4.
The total mass number before the decay is [tex]\( 210 \)[/tex].
The total mass number after the decay is [tex]\( 206 + 4 = 210 \)[/tex], which confirms the conservation of mass number.
Given that a helium nucleus ([tex]\(^{4}_{2}\text{He}\)[/tex]) is emitted from the parent nucleus, this specific type of radioactive decay is known as alpha decay.
Thus, the type of radioactive decay resulting from the transformation of [tex]\(^{210}_{84}\text{Po}\)[/tex] to [tex]\(^{206}_{82}\text{Pb}\)[/tex] with the emission of [tex]\(^{4}_{2}\text{He}\)[/tex] is:
C. alpha
