A uranium nucleus `""_(92)U^(238)` emits an `alpha`-particle and a `beta`-particle in succession. The atomic number and mass number of the final nucleus will be

Explanation

When an `alpha`- particle is emitted, we have the product nucleues `X`:
`._(92)U^(238) rarr ._(90)X^(234) +._2 He^(4)`
When the product nucleus emits a `beta-` particles, the final nucleus is `Y` (say) given by `._(90)X^(234) rarr ._(91)Y^(234) +._(-1)beta^(2)`
Thus, atomic number of final nucleus` =91`, mass number of final nucleus`=234`.