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

Explanation

To solve the problem of determining the atomic number and mass number of the final nucleus after a uranium nucleus \(_{92}^{238}\text{U}\) emits an alpha particle and a beta particle, we can follow these steps:

### Step 1: Identify the Initial Nucleus
The initial nucleus is uranium-238, denoted as \(_{92}^{238}\text{U}\). Here, the atomic number (Z) is 92, and the mass number (A) is 238.

### Step 2: Emission of an Alpha Particle
An alpha particle is essentially a helium nucleus, represented as \(_{2}^{4}\text{He}\). When an alpha particle is emitted, the changes in the atomic number and mass number of the uranium nucleus can be calculated as follows:

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