Examkin

Question

Difficulty: MediumNatural Radioactivity and Radiation Emissions

During a natural radioactive decay series, a parent nucleus of Actinium-227 (89227Ac^{227}_{89}\text{Ac}) emits 55 α\alpha-particles and 33 β−\beta^--particles to reach a stable state. What is the atomic number (ZZ) of the resulting stable daughter nucleus?

Answer: 82

Answer

The atomic number of the resulting stable daughter nucleus is 82.
Each alpha particle emission decreases the nuclear charge (atomic number) by 2, so emitting 5 alpha particles reduces the atomic number by 10. Each beta-minus emission increases the nuclear charge by 1, so emitting 3 beta-minus particles increases the atomic number by 3. Starting with an initial atomic number of 89, the resulting atomic number is 89 - 10 + 3 = 82.

Step-by-Step Solution

1
Determine the atomic number reduction from alpha emissions.
5 alpha particles reduce the atomic number by 5 * 2 = 10.
An alpha particle (^4_2He) carries 2 protons, so each emission decreases Z by 2.
2
Determine the atomic number increase from beta-minus emissions.
3 beta-minus particles increase the atomic number by 3 * 1 = 3.
A beta-minus particle (^0_-1e) is emitted when a neutron converts to a proton, increasing Z by 1.
3
Apply the conservation of atomic number to find the final value.
Z_final = 89 - 10 + 3 = 82.
Subtracting the alpha contribution and adding the beta-minus contribution from the initial atomic number gives the daughter nucleus's atomic number.

Key Concept

Conservation of atomic number (charge) in natural radioactive decay chains
Rate this question

Topics

Natural Radioactivity and Radiation Emissions

More questions from this topic