Question

Express pb -211 as a nuclear equation.

Answer 1

`\[ \large \text{Pb}^(211) \rightarrow \text{Tl}^(207) + \text{He}^(4) \]`

Step-by-step explanation:

`Lead-211 (\( \text{Pb}^(211) \))` undergoes alpha decay, resulting in
`thallium-207 (\( \text{Tl}^(207) \))` and a
`helium-4 (\( \text{He}^(4) \))` nucleus. Alpha decay involves the emission of an alpha particle, which consists of two protons and two neutrons. This process reduces the atomic number of the parent nucleus by 2 and the mass number by 4. In this case, lead-211 with an atomic number of 82 and a mass number of 211 transforms into thallium-207 with an atomic number of 81 and a mass number of 207. The helium-4 nucleus, consisting of two protons and two neutrons, is emitted as a byproduct.

Alpha decay is a common mode of radioactive decay for heavy nuclei, providing a stable configuration by reducing the excessive internal energy associated with the parent nucleus. In nuclear equations, the arrow represents the decay process, and the numbers above and below the chemical symbols indicate the atomic and mass numbers, respectively. This equation succinctly describes the transformation of lead-211 into thallium-207 and helium-4 through alpha decay. The conservation of mass and charge is evident, as the sum of the atomic numbers and mass numbers on both sides of the equation remains constant, reflecting the fundamental principles governing nuclear reactions.