Physics Nuclear Physics MDCAT 2015
PMDC Verified Question 69 of 98
In the reaction $$ _{90}^{234}\text{Th} \rightarrow _{91}^{234}\text{Pa} + \text{e} $$, the electron \(\text{e}\) emits from the
A
\(1^{\text{st}}\) orbit
B
Nucleus
C
\(2^{\text{nd}}\) orbit
D
Valence shell
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Propolis Cognitive Error Autopsy

Official Correct Choice:
Option B: Nucleus


Concept:

This equation represents a nuclear beta minus (\(\beta^{-}\)) decay, where Thorium-234 transmutes into Protactinium-234.

Solution:

  • Beta decay is strictly a nuclear process mediated by the weak force.


  • A neutron deep inside the nucleus transforms into a proton, creating an electron and an antineutrino in the process.


  • This highly energetic electron is immediately ejected directly from the nucleus, not from the atomic orbital electron clouds.


Why other options are incorrect:

Options A, C, and D refer to atomic orbitals (electron shells). Electrons ejected from there constitute ionization (via photoelectric effect or Auger effect), not nuclear transmutation.

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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