Physics Nuclear Physics MDCAT 2014
PMDC Verified Question 75 of 98
A uranium isotope \( _{92}^{234}\text{U} \) undergoes one \(\alpha\)-decay and one \(\beta\)-decay. What is the atomic number of the final product?
A
90
B
89
C
91
D
88
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option C: 91


Concept:

We must trace the sequential changes to the atomic number (\(Z\)) caused by each specific decay event.

Solution:

  • Step 1: Alpha Decay
    Emitting an alpha particle (\(_{2}^{4}\text{He}\)) subtracts 2 from the atomic number.
    $$ Z_{\text{intermediate}} = 92 - 2 = 90 $$


  • Step 2: Beta Decay
    Emitting a beta particle (\(_{-1}^{0}\text{e}\)) adds 1 to the atomic number.
    $$ Z_{\text{final}} = 90 + 1 = 91 $$


Why other options are incorrect:

90 is the state after just the alpha decay. 89 would require beta-plus or electron capture. 88 would be the result of two consecutive alpha decays.

Quality & Fidelity Assurance: Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.

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