A nucleus ${^{238}_{92}X}$ emits one alpha particle and then one beta particle (electron). The product nucleus will have

Atomic number 91, Mass number = 234

The correct answer is Option (2) → Atomic number 91, Mass number = 234

Parent nucleus: $_{92}^{238}X$

After $\alpha$-decay: $\ ^4_2He$ is emitted

$\ ^{238}_{92}X \;\;\to\;\; ^{234}_{90}Y$

So, $A=234$, $Z=90$

Now $\beta^-$-decay: one electron is emitted

$\ ^{234}_{90}Y \;\;\to\;\; ^{234}_{91}Z$

So, $A=234$, $Z=91$

Final Answer: Atomic number = 91, Mass number = 234