Consider the following reaction: `""_(7)^(14)N(d, alpha) ""_(6)^(12)C` where `""_(7)^(14)N` has a mass of 14.003u, `""_(6)^(12)C` has a mass of 12.000000u, d has a mass of 2.014 102 u and `alpha` has a mass of 4.002 603 u. How much energy is released in this reaction

Options

A. 1.2 MeV
B. 13.6 MeV
C. 8.3 MeV
D. 27.4 MeV

Answer

To find the energy released in the reaction \( _{7}^{14}N(d, \alpha) _{6}^{12}C \), we can follow these steps:

### Step 1: Identify the masses of the reactants and products
- Mass of \( _{7}^{14}N \) (Nitrogen) = 14.003 u
- Mass of \( d \) (Deuteron) = 2.014102 u
- Mass of \( _{6}^{12}C \) (Carbon) = 12.000000 u
- Mass of \( \alpha \) (Alpha particle) = 4.002603 u

### Step 2: Write the equation for the energy released
The energy released \( Q \) in the reaction can be calculated using the mass-energy equivalence principle, given by the formula:
\[ Q = (m_{\text{reactants}} - m_{\text{products}})c^2 \]
Where:
- \( m_{\text{reactants}} = m_d + m_{N} \)
- \( m_{\text{products}} = m_{C} + m_{\alpha} \)

### Step 3: Calculate the total mass of the reactants
\[ m_{\text{reactants}} = m_d + m_{N} = 2.014102 \, \text{u} + 14.003 \, \text{u} = 16.017102 \, \text{u} \]

### Step 4: Calculate the total mass of the products
\[ m_{\text{products}} = m_{C} + m_{\alpha} = 12.000000 \, \text{u} + 4.002603 \, \text{u} = 16.002603 \, \text{u} \]

### Step 5: Calculate the mass difference
\[ \Delta m = m_{\text{reactants}} - m_{\text{products}} = 16.017102 \, \text{u} - 16.002603 \, \text{u} = 0.014499 \, \text{u} \]

### Step 6: Convert the mass difference to energy
Using the conversion factor \( 1 \, \text{u} = 931.5 \, \text{MeV}/c^2 \):
\[ Q = \Delta m \cdot 931.5 \, \text{MeV/u} = 0.014499 \, \text{u} \cdot 931.5 \, \text{MeV/u} \approx 13.5 \, \text{MeV} \]

### Step 7: Final answer
The energy released in the reaction is approximately **13.5 MeV**.

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