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When ₉₂U²³⁸ decays it emits an $α$ -particle. The new nuclide in turn emits a $β$ -particle to give another nuclide X, The mass number and atomic number of X are, respectively

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234 and 91

234 and 96

232 and 88

234 and 88

answer is A.

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Detailed Solution

$_{92} U^{238} ⟶_{z} X^{A} +_{2} {He}^{4} +_{- 1} e^{0}$

Equating mass number on both sides

238 = A+4+0
.'. A = 238 - 4 = 234

Equating atomic number on both sides

92 = Z + 2 - 1
:. Z = 92- 1 = 91

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