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 process \( _{11}^{22}\text{Na} \to _{10}^{22}\text{Ne} + e^+ + \nu \), we will follow these steps:

### Step 1: Identify the masses involved
The Q-value of a nuclear reaction is calculated using the masses of the reactants and products. The mass of the reactant is the mass of sodium-22 (\( _{11}^{22}\text{Na} \)), and the products are neon-22 (\( _{10}^{22}\text{Ne} \)), a positron (\( e^+ \)), and a neutrino (\( \nu \)).

### Step 2: Write the equation for Q-value
The Q-value can be expressed as:
\[ Q = (m_{\text{reactants}} - m_{\text{products}}) c^2 \]
Where:
- \( m_{\text{reactants}} \) is the mass of sodium-22
- \( m_{\text{products}} \) is the sum of the masses of neon-22, the positron, and the neutrino

### Step 3: Insert the known masses
The known masses (in atomic mass units, u) are approximately:
- Mass of \( _{11}^{22}\text{Na} \) = 22.989769 u
- Mass of \( _{10}^{22}\text{Ne} \) = 21.991385 u
- Mass of positron \( e^+ \) = 0.00054858 u
- Mass of neutrino \( \nu \) is negligible (approximately 0)

Now, we can calculate the mass of the products:
\[ m_{\text{products}} = m_{\text{Ne}} + m_{e^+} + m_{\nu} \approx 21.991385 + 0.00054858 + 0 \approx 21.99193358 \, \text{u} \]

### Step 4: Calculate the Q-value
Now we can substitute the masses into the Q-value equation:
\[ Q = (m_{\text{Na}} - m_{\text{products}}) c^2 \]
\[ Q = (22.989769 - 21.99193358) c^2 \]
\[ Q \approx 0.99783542 \, c^2 \]

### Step 5: Convert to energy
Using \( c^2 \) in MeV (where \( 1 \, \text{u} \approx 931.5 \, \text{MeV} \)):
\[ Q \approx 0.99783542 \times 931.5 \, \text{MeV} \approx 930.5 \, \text{MeV} \]

### Conclusion
Thus, the Q-value of the decay \( _{11}^{22}\text{Na} \to _{10}^{22}\text{Ne} + e^+ + \nu \) is approximately **930.5 MeV**.

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