`._(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)`.

Explanation

To solve the problem, we need to analyze the decay processes of uranium-238 (U-238) step by step.

### Step 1: Identify the initial parameters of U-238
- Atomic number (Z) of U-238 = 92
- Mass number (A) of U-238 = 238

### Step 2: Determine the result of alpha decay
When uranium-238 undergoes alpha decay, it emits an alpha particle (which is a helium nucleus with an atomic number of 2 and a mass number of 4). The decay can be represented as follows:
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