`Na^(23)` is more stable isotope of `Na`. Find out the process by which `._(11)Na^(24)` can undergo radioactive decay.

Options

A. `beta^(c-)-` emission
B. `alph-`emission
C. `beta^(o+)-`emission
D. `K` electron capture.

Explanation

To find out the process by which \( \text{Na}^{24} \) can undergo radioactive decay, we will follow these steps:

### Step 1: Identify the Isotope
We are given the isotope \( \text{Na}^{24} \) (Sodium with mass number 24). The atomic number of sodium (Na) is 11, which means it has 11 protons.

### Step 2: Calculate the Number of Neutrons
To find the number of neutrons in \( \text{Na}^{24} \), we use the formula:
\[ \text{Number of Neutrons} = \text{Mass Number} - \text{Atomic Number} \]
For \( \text{Na}^{24} \):
\[ \text{Number of Neutrons} = 24 - 11 = 13 \]

### Step 3: Calculate the Neutron-to-Proton Ratio
Next, we calculate the neutron-to-proton (N/P) ratio:
\[ \text{N/P Ratio} = \frac{\text{Number of Neutrons}}{\text{Number of Protons}} = \frac{13}{11} \approx 1.18 \]
Since the N/P ratio is greater than 1, this indicates that \( \text{Na}^{24} \) is unstable.

### Step 4: Determine the Decay Process
To stabilize the nucleus, the N/P ratio needs to be reduced. This can be achieved by converting one neutron into one proton. This process occurs through beta decay (specifically, beta emission).

### Step 5: Write the Beta Decay Equation
In beta decay, a neutron is converted into a proton and a beta particle (an electron). The equation for this process can be written as:
\[ \text{n} \rightarrow \text{p} + \beta^{-} \]
Where:
- \( \text{n} \) is a neutron,
- \( \text{p} \) is a proton,
- \( \beta^{-} \) is the emitted beta particle.

### Step 6: Write the Complete Decay Reaction
For \( \text{Na}^{24} \), the complete decay reaction can be expressed as:
\[ \text{Na}^{24} \rightarrow \text{Mg}^{24} + \beta^{-} \]
Here, \( \text{Mg}^{24} \) (Magnesium with mass number 24) is formed after the decay.

### Conclusion
Thus, the process by which \( \text{Na}^{24} \) undergoes radioactive decay is through beta emission, resulting in the formation of \( \text{Mg}^{24} \) and a beta particle.

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