Q -value of the decay `""_(11)^(22)Na to ""_(10)^(22)Ne+e^(+)+nu` is

Options

A. `[m(""_(11)^(22)Na)-m(""_(10)^(22)Ne)]c^2`
B. `[m(""_(11)^(22)Na)-m(""_(10)^(22)Ne)-m_e]c^2`
C. `[m(""_(11)^(22)Na)-m(""_(10)^(22)Ne)-2m_e]c^2`
D. `[m(""_(11)^(22)Na)-m(""_(10)^(22)Ne)-3m_e]c^2`

Answer

To find the Q-value of the decay reaction \( _{11}^{22}\text{Na} \to _{10}^{22}\text{Ne} + e^{+} + \nu \), we will follow these steps:

### Step 1: Identify the Masses of Reactants and Products
- The reactant is \( _{11}^{22}\text{Na} \) (Sodium-22).
- The products are \( _{10}^{22}\text{Ne} \) (Neon-22), a positron \( e^{+} \), and a neutrino \( \nu \).

### Step 2: Write the Q-value Formula
The Q-value of a nuclear reaction is given by the mass defect multiplied by \( c^2 \):
\[ Q = \Delta m \cdot c^2 \]
where \( \Delta m = m_{\text{reactants}} - m_{\text{products}} \).

### Step 3: Calculate the Mass of Reactants
- The mass of the reactant \( m_{\text{reactants}} \) is simply the mass of Sodium-22:
\[ m_{\text{reactants}} = m(^{22}_{11}\text{Na}) \]

### Step 4: Calculate the Mass of Products
- The mass of the products includes the mass of Neon-22, the mass of the positron, and the mass of the neutrino (which we will neglect due to its very small mass):
\[ m_{\text{products}} = m(^{22}_{10}\text{Ne}) + m(e^{+}) + m(\nu) \]
Since the mass of the neutrino is negligible, we can ignore it:
\[ m_{\text{products}} \approx m(^{22}_{10}\text{Ne}) + m(e^{+}) \]

### Step 5: Substitute into the Q-value Formula
Substituting the masses into the Q-value formula gives:
\[ Q = m(^{22}_{11}\text{Na}) - \left(m(^{22}_{10}\text{Ne}) + m(e^{+})\right) \cdot c^2 \]

### Step 6: Simplify the Expression
Since the mass of the positron \( m(e^{+}) \) is equal to the mass of an electron \( m(e^{-}) \), we can write:
\[ Q = m(^{22}_{11}\text{Na}) - m(^{22}_{10}\text{Ne}) - 2m(e^{-}) \cdot c^2 \]

### Step 7: Conclusion
The final expression for the Q-value of the decay is:
\[ Q = m(^{22}_{11}\text{Na}) - m(^{22}_{10}\text{Ne}) - 2m(e^{-}) \cdot c^2 \]
Thus, the Q-value of the decay \( _{11}^{22}\text{Na} \to _{10}^{22}\text{Ne} + e^{+} + \nu \) is \( -2m(e^{-}) \cdot c^2 \).

### Final Answer
The Q-value of the decay is \( -2m_e c^2 \), where \( m_e \) is the mass of an electron.

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