`._(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

Explanation

To solve the problem, we need to find the atomic number and mass number of the residual nucleus after an alpha emission and then the nucleus after a beta emission. Let's break it down step by step.

### Step 1: Identify the initial nucleus
The initial nucleus is Uranium-238, which can be represented as:
\[ _{92}^{238}\text{U} \]
Here, the atomic number (Z) is 92 and the mass number (A) is 238.

### Step 2: Determine the effect of alpha emission
An alpha particle consists of 2 protons and 2 neutrons, which means it has an atomic number of 2 and a mass number of 4. When Uranium-238 emits an alpha particle, the new nucleus (let's call it U_X1) will have:
- Atomic number:
\[ Z_{X1} = Z_{U} - 2 = 92 - 2 = 90 \]
- Mass number:
\[ A_{X1} = A_{U} - 4 = 238 - 4 = 234 \]

Thus, the residual nucleus U_X1 can be represented as:
\[ _{90}^{234}\text{U} \]

### Step 3: Determine the effect of beta emission
Beta emission involves the conversion of a neutron into a proton, which increases the atomic number by 1 while the mass number remains unchanged. For U_X1:
- Atomic number after beta emission:
\[ Z_{X2} = Z_{X1} + 1 = 90 + 1 = 91 \]
- Mass number remains the same:
\[ A_{X2} = A_{X1} = 234 \]

Thus, the nucleus U_X2 can be represented as:
\[ _{91}^{234}\text{U} \]

### Summary of Results
- For U_X1 (after alpha emission):
- Atomic number = 90
- Mass number = 234
- For U_X2 (after beta emission):
- Atomic number = 91
- Mass number = 234

### Final Answer
- U_X1: Atomic number = 90, Mass number = 234
- U_X2: Atomic number = 91, Mass number = 234

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