Question

A decay chain of the nucleus $^{238}_{92}\text{U}$ involves eight $\alpha$-decays and six $\beta$-decays. The final nucleus at the end of the process will be

• A. $Z = 76; A = 200$
• B. $Z = 84; A = 206$
• C. $Z = 84; A = 224$
• D. $Z = 82; A = 206$
• E. $Z = 82; A = 200$

Hint:

$\beta$-decays only affect the charge (atomic number), never the mass. Always solve for the change in mass number ($A$) using $\alpha$-decays first.

Answer 1

The Correct Option is D

Concept:
[itemsep=4pt]
• $\alpha$-decay: Mass number ($A$) decreases by 4, Atomic number ($Z$) decreases by 2.
• $\beta$-decay ($\beta^-$): Mass number ($A$) remains same, Atomic number ($Z$) increases by 1.

Step 1: Calculate final Mass Number ($A'$).
Initial $A = 238$. With 8 $\alpha$-decays: \[ A' = 238 - (8 \times 4) = 238 - 32 = 206 \]

Step 2: Calculate final Atomic Number ($Z'$).
Initial $Z = 92$. With 8 $\alpha$-decays and 6 $\beta$-decays: \[ Z' = 92 - (8 \times 2) + (6 \times 1) = 92 - 16 + 6 = 82 \] The nucleus with $Z = 82$ is Lead (Pb).