Which of the following `"is"//"are"` correct when a nuclide of mass number `(A)` and atomic number `(Z)` undergoes radioactive process?

Options

A. Both `A` and `Z` decreases, the process is called `alpha`-decay.
B. A remains unchanged and `Z` decreases by 1. The process is called `beta^(o+)` or positron decay of `K-`electron capture.
C. Both `A` and `Z` remain unchanged, the process is called `gamma`-decay.
D. Both `A` and `Z` increase, the process is called nuclear isomerism.

Answer

To determine which statements are correct when a nuclide of mass number \( A \) and atomic number \( Z \) undergoes a radioactive process, we will analyze each option provided in the question.

### Step-by-step Solution:

1. **Option A: Both A and Z decrease.**
- This statement refers to alpha decay. In alpha decay, an alpha particle (which consists of 2 protons and 2 neutrons, equivalent to a helium nucleus) is emitted from the nucleus.
- When an alpha particle is emitted:
- The mass number \( A \) decreases by 4 (since 2 protons and 2 neutrons are lost).
- The atomic number \( Z \) decreases by 2 (since 2 protons are lost).
- Therefore, this option is **correct**.

2. **Option B: A remains unchanged and Z decreases by 1.**
- This statement refers to beta positive decay (positron emission) or electron capture.
- In beta positive decay, a positron (which has a mass number of 0 and an atomic number of +1) is emitted.
- When a positron is emitted:
- The mass number \( A \) remains unchanged (no nucleons are lost).
- The atomic number \( Z \) decreases by 1 (since one proton is converted into a neutron).
- Therefore, this option is **correct**.

3. **Option C: Both A and Z remain unchanged.**
- This statement refers to gamma decay. In gamma decay, the nucleus transitions from a higher energy state to a lower energy state by emitting a gamma photon.
- During this process:
- The mass number \( A \) remains unchanged.
- The atomic number \( Z \) remains unchanged.
- Therefore, this option is **correct**.

4. **Option D: Both A and Z increase.**
- This statement refers to nuclear isomerism. In nuclear isomerism, the nucleus exists in a metastable state, but there is no emission of particles that would change the mass number or atomic number.
- Therefore, both \( A \) and \( Z \) remain unchanged.
- Thus, this option is **incorrect**.

### Conclusion:
The correct options are:
- **Option A**: Correct
- **Option B**: Correct
- **Option C**: Correct
- **Option D**: Incorrect