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Zorluk: OrtaNatural 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?

Cevap: 82

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

Conservation of atomic number (charge) in natural radioactive decay chains
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Konular

Natural Radioactivity and Radiation Emissions

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