Physics · Nuclear Physics

NEET (UG) 2024 — Question 20

${}^{290}_{82}X \xrightarrow{\alpha} Y \xrightarrow{e^+} Z \xrightarrow{\beta^-} P \xrightarrow{e^-} Q$

In the nuclear emission stated above, the mass number and atomic number of the product $Q$ respectively, are

Answer: D

The parent nucleus is $^{290}_{82}\mathrm{X}$ .

α-decay: mass number decreases by 4, atomic number by 2 → $^{286}_{80}\mathrm{Y}$ .
β⁺-decay (positron emission): mass number unchanged, atomic number decreases by 1 → $^{286}_{79}\mathrm{Z}$ .
β⁻-decay: mass number unchanged, atomic number increases by 1 → $^{286}_{80}\mathrm{P}$ .
Another β⁻-decay: mass number unchanged, atomic number increases by 1 → $^{286}_{81}\mathrm{Q}$ . Thus mass number = 286, atomic number = 81. Hence option D is correct.

Answer key and solution verified before publishing.

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