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

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The discussion centers on a radioactive nucleus with an approximate mass of 212 and 92 protons undergoing beta and alpha decay. After beta decay, the atomic number increases by one, resulting in 93 protons, followed by alpha decay, which decreases the atomic number by two and the mass by four, leading to a final nucleus with 91 protons. Participants clarify that Protactinium, not Uranium, has 91 protons, emphasizing the importance of understanding the decay process rather than the specific isotopes involved. The conversation highlights the complexities of radioactive decay and the need for accurate identification of elements. Ultimately, the focus remains on the final nucleus resulting from the described decay sequence.
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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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