Physics Nuclear Physics MDCAT 2013
PMDC Verified Question 79 of 98
Emission of alpha decay from a radioactive substance cause:
A
Decrease in 'Z' by 4 and decrease in 'A' by 2
B
Decrease in 'Z' by 1 and 'A' remains same
C
Decreases in 'A' by 1 and 'Z' remains same
D
Decrease in 'A' by 4 and decrease in 'Z' by 2
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option D: Decrease in 'A' by 4 and decrease in 'Z' by 2


Concept:

An alpha particle is a helium-4 nucleus, denoted as \(_{2}^{4}\text{He}\).

Formula:

$$ _{Z}^{A}\text{X} \rightarrow _{Z-2}^{A-4}\text{Y} + _{2}^{4}\text{He} $$

Solution:

  • The ejected particle removes 2 protons from the parent nucleus. Thus, the atomic number \(Z\) decreases by 2.


  • The ejected particle removes 4 total nucleons (2 protons + 2 neutrons). Thus, the mass number \(A\) decreases by 4.


Why other options are incorrect:

Option A perfectly swaps the properties of \(Z\) and \(A\). Option B describes positron emission or electron capture. Option C is physically non-standard.

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