A positron is emitted from `._(11)Na^(23)` . The ratio of the atomic mass and atomic number of the resulting nuclide is

Options

A. `22//10`
B. `22//11`
C. `23//10`
D. `23//12`

Explanation

To solve the problem, we need to analyze the emission of a positron from the nuclide \( _{11}^{23}\text{Na} \) (sodium-23). Here’s a step-by-step breakdown of the solution:

### Step 1: Understand Positron Emission
When a positron is emitted, a proton in the nucleus is transformed into a neutron and a positron. This process decreases the atomic number of the element by 1, while the atomic mass remains unchanged.

### Step 2: Identify the Initial Nuclide
The initial nuclide is \( _{11}^{23}\text{Na} \):
- Atomic number (Z) = 11 (this represents sodium)
- Atomic mass (A) = 23

### Step 3: Determine the Resulting Nuclide
After the emission of a positron:
- The atomic number decreases by 1:
\[ Z' = 11 - 1 = 10 \]
- The atomic mass remains the same:
\[ A' = 23 \]
Thus, the resulting nuclide is \( _{10}^{23}X \), where \( X \) is the new element with atomic number 10, which corresponds to neon (Ne).

### Step 4: Calculate the Ratio of Atomic Mass to Atomic Number
Now, we need to find the ratio of the atomic mass to the atomic number of the resulting nuclide:
\[ \text{Ratio} = \frac{\text{Atomic Mass}}{\text{Atomic Number}} = \frac{A'}{Z'} = \frac{23}{10} \]

### Step 5: Final Answer
The ratio of the atomic mass to the atomic number of the resulting nuclide is:
\[ \frac{23}{10} \]