A uranium nucleus (atomic number `92`, mass number `231`) emits an `alpha`-particle and the resultant nucleus emits a `beta`-particles. What are the atomic and mass numbers of the final nucleus? .

Options

A. `90,233`
B. `90,238`
C. `91,234
D. `93,238`

Explanation

To solve the problem, we need to analyze the nuclear reactions step by step.

### Step 1: Identify the initial nucleus
The initial nucleus is a uranium nucleus with an atomic number (Z) of 92 and a mass number (A) of 231. We can represent this as:
\[ \text{Uranium} \, (^{231}_{92}\text{U}) \]

### Step 2: Emission of an alpha particle
When an alpha particle is emitted, the atomic number decreases by 2 and the mass number decreases by 4. An alpha particle is represented as:
\[ \text{Alpha particle} \, (^{4}_{2}\text{He}) \]

So, after the emission of an alpha particle, we can calculate the new atomic and mass numbers:
- New atomic number (Z) = 92 - 2 = 90
- New mass number (A) = 231 - 4 = 227

The resultant nucleus after alpha decay is:
\[ \text{New nucleus} \, (^{227}_{90}\text{X}) \]
where X is the new element with atomic number 90 (which is Thorium, Th).

### Step 3: Emission of a beta particle
Next, the resultant nucleus emits a beta particle. A beta particle emission increases the atomic number by 1 while the mass number remains unchanged. A beta particle is represented as:
\[ \text{Beta particle} \, (^{0}_{-1}\text{e}) \]

So, after the emission of a beta particle, we can calculate the new atomic and mass numbers:
- New atomic number (Z) = 90 + 1 = 91
- Mass number (A) remains the same = 227

The final nucleus after beta decay is:
\[ \text{Final nucleus} \, (^{227}_{91}\text{Y}) \]
where Y is the new element with atomic number 91 (which is Protactinium, Pa).

### Conclusion
The atomic number of the final nucleus is 91, and the mass number is 227.

### Final Answer
- Atomic number = 91
- Mass number = 227

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