A nuclear reaction given by
`1_Z X^A rarr .(Z+1) Y^A + ._(-1)e^0 + vecp` represents.

Options

A. fusion
B. fission
C. `beta`-decay
D. `gamma`-decay

Answer

To analyze the nuclear reaction given by:

\[ {}^{A}_{Z}X \rightarrow {}^{A}_{Z+1}Y + \bar{\nu} + e^{-} \]

we will break it down step by step.

### Step 1: Identify the Components of the Reaction
The reaction involves:
- A nucleus \(X\) with mass number \(A\) and atomic number \(Z\).
- A product nucleus \(Y\) with the same mass number \(A\) but an increased atomic number \(Z + 1\).
- An emitted electron (denoted as \(e^{-}\)).
- An emitted anti-neutrino (denoted as \(\bar{\nu}\)).

### Step 2: Analyze the Change in Atomic Number
In this reaction, the atomic number of the nucleus increases from \(Z\) to \(Z + 1\). This indicates that a proton is being added to the nucleus, which is characteristic of a beta decay process.

### Step 3: Identify the Type of Decay
There are two types of beta decay:
- **Beta-minus decay**: In this process, a neutron is converted into a proton, emitting an electron and an anti-neutrino. The atomic number increases by 1 while the mass number remains the same.
- **Beta-plus decay**: In this process, a proton is converted into a neutron, emitting a positron and a neutrino. The atomic number decreases by 1.

Given that the atomic number increases by 1 in our reaction, it indicates that this is a **beta-minus decay**.

### Step 4: Conclusion
Thus, the nuclear reaction can be concluded as a **beta-minus decay** where a neutron in nucleus \(X\) is transformed into a proton, resulting in nucleus \(Y\) and emitting an electron and an anti-neutrino.

### Final Answer
The reaction represents a **beta-minus decay**.

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