Find the Q - value for the nuclear reaction:
`._(2)He^(4)+._(7)N^(14) rarr ._(8)O^(17)+._(1)H^(1)`
`m(._(2)He^(4))=4.0039 "amu" m(._(7) N^(14))=14.0075 "amu" m(._(8)O^(17))=17.0045` amu
`m(._(1)H^(1))=1.0082` amu

Options

A. `2.54` MeV
B. `1.54` MeV
C. `5` MeV
D. `8` MeV

Answer

To find the Q-value for the nuclear reaction:

\[ \text{He}^{4}_{2} + \text{N}^{14}_{7} \rightarrow \text{O}^{17}_{8} + \text{H}^{1}_{1} \]

we will follow these steps:

### Step 1: Identify the masses of the reactants and products
- Mass of Helium (\( \text{He}^{4}_{2} \)): \( m(\text{He}) = 4.0039 \, \text{amu} \)
- Mass of Nitrogen (\( \text{N}^{14}_{7} \)): \( m(\text{N}) = 14.0075 \, \text{amu} \)
- Mass of Oxygen (\( \text{O}^{17}_{8} \)): \( m(\text{O}) = 17.0045 \, \text{amu} \)
- Mass of Hydrogen (\( \text{H}^{1}_{1} \)): \( m(\text{H}) = 1.0082 \, \text{amu} \)

### Step 2: Calculate the total mass of the reactants
The total mass of the reactants is the sum of the masses of Helium and Nitrogen:

\[ m_{\text{reactants}} = m(\text{He}) + m(\text{N}) = 4.0039 \, \text{amu} + 14.0075 \, \text{amu} \]

Calculating this gives:

\[ m_{\text{reactants}} = 18.0114 \, \text{amu} \]

### Step 3: Calculate the total mass of the products
The total mass of the products is the sum of the masses of Oxygen and Hydrogen:

\[ m_{\text{products}} = m(\text{O}) + m(\text{H}) = 17.0045 \, \text{amu} + 1.0082 \, \text{amu} \]

Calculating this gives:

\[ m_{\text{products}} = 18.0127 \, \text{amu} \]

### Step 4: Calculate the mass defect
The mass defect (\( \Delta m \)) is the difference between the total mass of the reactants and the total mass of the products:

\[ \Delta m = m_{\text{reactants}} - m_{\text{products}} = 18.0114 \, \text{amu} - 18.0127 \, \text{amu} \]

Calculating this gives:

\[ \Delta m = -0.0013 \, \text{amu} \]

### Step 5: Calculate the Q-value
The Q-value can be calculated using the formula:

\[ Q = \Delta m \times 931 \, \text{MeV} \]

Substituting the mass defect:

\[ Q = -0.0013 \, \text{amu} \times 931 \, \text{MeV} \]

Calculating this gives:

\[ Q \approx -1.21 \, \text{MeV} \]

### Conclusion
The Q-value for the nuclear reaction is approximately \( -1.21 \, \text{MeV} \).

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