`._(92)U^(238)` is a natural `alpha`-emitter. After `alpha`-emission the residual nucleus `U_(X1)` in turn emits a `beta`-particle to produce another nucleus `U_(X2)`. Find out the atomic number and mass number of `U_(X1)` and `U_(X2)`.

Options

A. 234 and 91
B. 234 and 96
C. 232 and 88
D. 234 and 88

Answer

A

Explanation

`._(92)U^(238) rarr ._(2)X^(A) + ._(2)He^(4) + ._(-1)e^(0)`
Equation 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`