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

Answer

To find the Q-value for the nuclear reaction:

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

we will use the formula for Q-value:

\[ Q = \text{(mass of reactants)} - \text{(mass of products)} \]

### Step 1: Calculate the total mass of the reactants

The reactants in this reaction are Helium-4 and Nitrogen-14.

- Mass of Helium-4, \( m(\text{ }_{2}\text{He}^{4}) = 4.0039 \, \text{amu} \)
- Mass of Nitrogen-14, \( m(\text{ }_{7}\text{N}^{14}) = 14.0075 \, \text{amu} \)

Now, we sum the masses of the reactants:

\[ \text{Total mass of reactants} = m(\text{ }_{2}\text{He}^{4}) + m(\text{ }_{7}\text{N}^{14}) = 4.0039 \, \text{amu} + 14.0075 \, \text{amu} \]

\[ \text{Total mass of reactants} = 18.0114 \, \text{amu} \]

### Step 2: Calculate the total mass of the products

The products in this reaction are Oxygen-17 and Hydrogen-1.

- Mass of Oxygen-17, \( m(\text{ }_{8}\text{O}^{17}) = 17.0045 \, \text{amu} \)
- Mass of Hydrogen-1, \( m(\text{ }_{1}\text{H}^{1}) = 1.0082 \, \text{amu} \)

Now, we sum the masses of the products:

\[ \text{Total mass of products} = m(\text{ }_{8}\text{O}^{17}) + m(\text{ }_{1}\text{H}^{1}) = 17.0045 \, \text{amu} + 1.0082 \, \text{amu} \]

\[ \text{Total mass of products} = 18.0127 \, \text{amu} \]

### Step 3: Calculate the Q-value

Now we can calculate the Q-value using the masses we found:

\[ Q = \text{(mass of reactants)} - \text{(mass of products)} \]

\[ Q = 18.0114 \, \text{amu} - 18.0127 \, \text{amu} \]

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

### Conclusion

The Q-value for the nuclear reaction is:

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

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