What is the final nucleus after a series of radioactive decays?

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The discussion revolves around a radioactive nucleus with an approximate mass of 212 and 92 protons undergoing a series of decays, specifically beta decay followed by alpha decay. Participants are exploring the characteristics of the final nucleus resulting from these decay processes.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the effects of beta and alpha decay on the atomic number and mass of the nucleus, questioning the feasibility of the final nucleus with 91 protons. There is also a consideration of the isotopes involved in the decay process.

Discussion Status

The discussion is ongoing, with participants providing insights into the decay processes and questioning the existence of certain isotopes. Some guidance has been offered regarding the changes in atomic number and mass during the decays, but no consensus has been reached on the final nucleus.

Contextual Notes

There is a mention of constraints regarding the isotopes of Uranium and Protactinium, indicating that the existence of certain isotopes may not be critical to the problem at hand.

athenaroa
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modern physics

A certain radioactive nucleus has ' approximate mass' 212 and contains 92 protons. The nucleus undergoes beta decay. The remaining (product) nucleus, still radioactive, undergoes alpha decay, leaving yet a new nucleus. The number of protons in this 'final' new nucleus is:
 
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beta decay the atomic number is raised by one, alpha decay the relative atomic mass decreases by 4 and the atomic number decreases by 2.
 
91. However, there is no such beast. The lightest isotope of Uranium has 218 nucleons, and it alpha decays.
 
mathman said:
91. However, there is no such beast. The lightest isotope of Uranium has 218 nucleons, and it alpha decays.

Uranium doesn't have 91 protons. Protactinium does. I don't think the existence of the isotope is necessarily an important part of the exercise.
 

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