PMDC Verified
Question 77 of 98
A radioactive isotope W decays to X which decays to Y and Y decays to Z as represented by the figure below:
What is the change in the atomic number from W to Z?
Propolis Cognitive Error Autopsy
Official Correct Choice:
Option C: Decreases by 3
Concept:Track the sequential algebraic changes to the atomic number (\(Z\)) throughout the nuclear decay chain.
Rules of Radioactive Decay:- \(\beta^-\) Decay: A neutron converts into a proton, emitting an electron and antineutrino. Atomic number increases by 1: \(\Delta Z = +1\).
- \(\alpha\) Decay: Emission of a helium nucleus (\(^4_2\text{He}\)). Atomic number decreases by 2: \(\Delta Z = -2\).
Solution:- Let the initial atomic number of isotope W be \(Z\).
- Step 1 (\(\beta^-\) decay): \(\text{W} \rightarrow \text{X}\)
$$Z_X = Z + 1$$
- Step 2 (\(\alpha\) decay): \(\text{X} \rightarrow \text{Y}\)
$$Z_Y = (Z + 1) - 2 = Z - 1$$
- Step 3 (\(\alpha\) decay): \(\text{Y} \rightarrow \text{Z}\)
$$Z_Z = (Z - 1) - 2 = Z - 3$$
- Total Change (\(\Delta Z_{\text{net}}\)):
$$\Delta Z = Z_Z - Z_W = (Z - 3) - Z = -3$$
- Therefore, the atomic number decreases by 3 (Option C).
Why other options are incorrect:- Option A (Increases by 3): Inverts sign by erroneously treating alpha decay as an addition to atomic number.
- Option B & Option D: Arithmetic miscalculation of nucleon vs. proton losses.
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