Consider the following decay `._(Z)^(A)Xrarr_(Z+1)^(A)Y+_(-1)^(0)e,X` is unstable because:

Options

A. its nucleus has excess energy
B. `(n)/(p)` ratio is high
C. `(n)/(p)` ratio is low
D. none of these

Answer

To understand why the decay process \( _{Z}^{A}X \rightarrow _{Z+1}^{A}Y + _{-1}^{0}e \) indicates that the nucleus \( X \) is unstable, we can analyze the situation step by step.

### Step-by-Step Solution:

1. **Identify the Decay Process**:
The decay process given is \( _{Z}^{A}X \rightarrow _{Z+1}^{A}Y + _{-1}^{0}e \). Here, \( X \) is a nucleus with atomic number \( Z \) and mass number \( A \) that decays into a nucleus \( Y \) with atomic number \( Z+1 \) and the emission of a beta particle \( _{-1}^{0}e \) (an electron).

2. **Understanding Atomic Number and Mass Number**:
- The atomic number \( Z \) represents the number of protons in the nucleus.
- The mass number \( A \) represents the total number of protons and neutrons in the nucleus.

3. **Beta Decay Mechanism**:
In beta decay, a neutron in the nucleus is transformed into a proton, resulting in an increase in the atomic number by 1 (from \( Z \) to \( Z+1 \)), while the mass number remains the same (remains \( A \)). This process can be represented as:
\[ n \rightarrow p + e^{-} + \bar{\nu} \]
where \( n \) is a neutron, \( p \) is a proton, \( e^{-} \) is the emitted beta particle (electron), and \( \bar{\nu} \) is an antineutrino.

4. **Neutron-to-Proton Ratio (n/p Ratio)**:
- The stability of a nucleus is often determined by the ratio of neutrons to protons (n/p ratio).
- For stable nuclei, this ratio is typically around 1 for lighter elements, but as the atomic number increases, the ratio tends to increase (more neutrons are needed to stabilize the nucleus).

5. **Analyzing the Stability of Nucleus \( X \)**:
- Since the decay process results in an increase in the atomic number (from \( Z \) to \( Z+1 \)), it implies that the number of protons is increasing.
- If the n/p ratio is high (more neutrons than protons), the nucleus is likely unstable and will undergo decay to reach a more stable state.

6. **Conclusion**:
The nucleus \( X \) is unstable because it has a high n/p ratio, which leads to an excess of neutrons compared to protons. This instability drives the decay process, resulting in the emission of a beta particle and the transformation into a more stable nucleus \( Y \).