During alpha-decay , a nucleus decays by emitting an `alpha`-particle ( a helium nucleus `._2He^4`) according to the equation
`._Z^AX to ._(Z-2)^(A-4)Y+._2^4He+Q`
In this process, the energy released Q is shared by the emitted `alpha`-particle and daughter nucleus in the form of kinetic energy .
The energy Q is divided in a definite ratio among the `alpha`-particle and the daughter nucleus .
A nucleus that decays spontaneously by emitting an electron or a positron is said to undergo `beta`-decay .This process also involves a release of definite energy . Initially, the `beta`-decay was represented as `._Z^AX to ._(Z+1)^AY + e^(-)"(electron)"+Q`
According to this reaction, the energy released during each decay must be divided in definite ratio by the emitted e' (`beta`-particle) and the daughter nucleus. While , in alpha decay, it has been found that every emitted `alpha`-particle has the same sharply defined kinetic energy. It is not so in case of `beta`-decay . The energy of emitted electrons or positrons is found to vary between zero to a certain maximum value. Wolfgang Pauli first suggested the existence of neutrinoes in 1930. He suggested that during `beta`-decay, a third particle is also emitted. It shares energy with the emitted `beta` particles and thus accounts for the energy distribution.
During `beta^+` decay (positron emission) a proton in the nucleus is converted into a neutron, positron and neutrino. The reaction is correctly represented as

Options

A. `._Z^AX to ._(Z-1)^(A-1)Y + e^(+) + v`+ Energy
B. `._Z^AX to ._(Z-1)^(" "A) X + e^(+) + v`+ Energy
C. `._Z^AX to ._(Z-1)^A Y + e^(+) + v`+ Energy
D. `._Z^AX to ._Z^(A-1)X + e^(+) +v +` Energy

Answer

To solve the problem regarding beta decay (positron emission), we need to analyze the process step by step.

### Step-by-Step Solution:

1. **Understanding Beta Decay**: In beta decay, specifically beta-plus decay, a proton in the nucleus is transformed into a neutron, emitting a positron and a neutrino. This process can be represented by the equation:
\[ _Z^A X \rightarrow _{Z-1}^A Y + e^+ + \nu + Q \]
where \( _Z^A X \) is the parent nucleus, \( _{Z-1}^A Y \) is the daughter nucleus, \( e^+ \) is the emitted positron, \( \nu \) is the neutrino, and \( Q \) is the energy released.

2. **Identifying Changes in Atomic Number and Mass Number**:
- The atomic number \( Z \) decreases by 1 because one proton is converted into a neutron.
- The mass number \( A \) remains unchanged because the total number of nucleons (protons + neutrons) stays the same.

3. **Writing the Reaction**:
- The initial nucleus has \( Z \) protons and \( A \) nucleons.
- After the decay, the new nucleus has \( Z-1 \) protons and still \( A \) nucleons, which means it has \( A - (Z-1) = A - Z + 1 \) neutrons.
- Therefore, the reaction can be summarized as:
\[ _Z^A X \rightarrow _{Z-1}^A Y + e^+ + \nu + Q \]

4. **Finalizing the Equation**:
- The complete equation for beta-plus decay is:
\[ _Z^A X \rightarrow _{Z-1}^A Y + e^+ + \nu + Q \]
This indicates that the original nucleus \( X \) transforms into a new nucleus \( Y \) with one less proton, while emitting a positron and a neutrino along with the energy \( Q \).

5. **Conclusion**: The correct representation of the beta-plus decay process is confirmed as:
\[ _Z^A X \rightarrow _{Z-1}^A Y + e^+ + \nu + Q \]