Find the binding energy of `Na^(23)` . Atomic mass of `Na^(23)` is 22.9898 amu and that of ,,,,,, is 1.00783 amu. The mass of neutron = 1.00867 amu.

Options

A. 931 MeV
B. 186.54 MeV
C. 5.38 MeV
D. None of these

Explanation

To find the binding energy of \( \text{Na}^{23} \), we will follow these steps:

### Step 1: Identify the components of the sodium nucleus
- Sodium (\( \text{Na}^{23} \)) has:
- Number of protons (\( Z \)) = 11
- Number of neutrons (\( N \)) = 12 (since \( 23 - 11 = 12 \))
- Number of electrons = 11 (equal to the number of protons)

### Step 2: Calculate the mass of the nucleus
- The mass of the nucleus can be calculated using the formula:
\[ M = m - (Z \times m_e) \]
where:
- \( M \) = mass of the nucleus
- \( m \) = atomic mass of \( \text{Na}^{23} = 22.9898 \, \text{amu} \)
- \( Z \) = number of protons = 11
- \( m_e \) = mass of an electron (approximately \( 0.00054858 \, \text{amu} \))

Calculating the mass of the nucleus:
\[ M = 22.9898 \, \text{amu} - (11 \times 0.00054858 \, \text{amu}) = 22.9898 \, \text{amu} - 0.00603438 \, \text{amu} \approx 22.983766 \, \text{amu} \]

### Step 3: Calculate the mass defect
- The mass defect (\( \Delta m \)) is given by:
\[ \Delta m = (N \times m_n + Z \times m_p) - M \]
where:
- \( m_n \) = mass of a neutron = \( 1.00867 \, \text{amu} \)
- \( m_p \) = mass of a proton = \( 1.00783 \, \text{amu} \)

Calculating the mass defect:
\[ \Delta m = (12 \times 1.00867 \, \text{amu} + 11 \times 1.00783 \, \text{amu}) - 22.983766 \, \text{amu} \]
Calculating the individual masses:
\[ = (12.10404 \, \text{amu} + 11.08613 \, \text{amu}) - 22.983766 \, \text{amu} \]
\[ = 23.19017 \, \text{amu} - 22.983766 \, \text{amu} \approx 0.206404 \, \text{amu} \]

### Step 4: Calculate the binding energy
- The binding energy (\( E_b \)) can be calculated using the formula:
\[ E_b = \Delta m \times 931 \, \text{MeV} \]
Substituting the value of \( \Delta m \):
\[ E_b = 0.206404 \, \text{amu} \times 931 \, \text{MeV/amu} \approx 192.4 \, \text{MeV} \]

### Final Answer
The binding energy of \( \text{Na}^{23} \) is approximately **192.4 MeV**.

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