Calculate the energy of the following nuclear reaction:
`._(1)H^(2)+._(1)H^(3) to ._(2)He^(4) + ._(0)n^(1)+Q`
Given: `m(._(1)H^(2))=2.014102u, m(._(1)H^(3))=3.016049u, m(._(2)He^(4))=4.002603u, m(._(0)n^(1))=1.008665u`

Explanation

To calculate the energy of the nuclear reaction given, we will follow these steps:

### Step 1: Identify the masses of the reactants and products.
The nuclear reaction is:
\[ _{1}^{2}H + _{1}^{3}H \rightarrow _{2}^{4}He + _{0}^{1}n + Q \]

Given masses:
- Mass of \( _{1}^{2}H \) (Deuterium) = \( 2.014102 \, u \)
- Mass of \( _{1}^{3}H \) (Tritium) = \( 3.016049 \, u \)
- Mass of \( _{2}^{4}He \) (Helium-4) = \( 4.002603 \, u \)
- Mass of \( _{0}^{1}n \) (Neutron) = \( 1.008665 \, u \)

### Step 2: Calculate the total mass of the reactants.
The total mass of the reactants is the sum of the masses of Deuterium and Tritium:
\[ \text{Mass of reactants} = m( _{1}^{2}H ) + m( _{1}^{3}H ) = 2.014102 \, u + 3.016049 \, u \]
\[ \text{Mass of reactants} = 5.030151 \, u \]

### Step 3: Calculate the total mass of the products.
The total mass of the products is the sum of the masses of Helium-4 and the neutron:
\[ \text{Mass of products} = m( _{2}^{4}He ) + m( _{0}^{1}n ) = 4.002603 \, u + 1.008665 \, u \]
\[ \text{Mass of products} = 5.011268 \, u \]

### Step 4: Calculate the mass defect (\( \Delta m \)).
The mass defect is the difference between the total mass of the reactants and the total mass of the products:
\[ \Delta m = \text{Mass of reactants} - \text{Mass of products} \]
\[ \Delta m = 5.030151 \, u - 5.011268 \, u = 0.018883 \, u \]

### Step 5: Convert the mass defect to energy.
The energy released in the reaction (\( Q \)) can be calculated using the formula:
\[ Q = \Delta m \times 931 \, \text{MeV/u} \]
Substituting the value of \( \Delta m \):
\[ Q = 0.018883 \, u \times 931 \, \text{MeV/u} \approx 17.58 \, \text{MeV} \]

### Final Answer:
The energy released in the nuclear reaction is approximately \( 17.58 \, \text{MeV} \).

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