The source of energy of stars is nuclear fusion. Fusion reaction occurs at very high temperature, about `10^(7) `. Energy released in the process of fusion is due to mass defect. It is also called `Q`-value. `Q = Delta mc^(2), Delta m =` mass defect.
In a nuclear reaction
`._(1)H^(2) + ._(1)H^(2) rarr ._(2)He^(3) + ._(0)n^(1)`
If the masses of ._(1)H^(2)` and `._(2)He^(3)` are 2.014741 and 3.016977 amu, respectively. then the `Q`-value of the reaction is nearly.

Options

A. `0.00352 MeV`
B. `3.27 MeV`
C. `0.82 MeV`
D. `2.45 MeV`

Explanation

To solve the problem, we need to calculate the Q-value of the nuclear fusion reaction given by:

\[ _{1}H^{2} + _{1}H^{2} \rightarrow _{2}He^{3} + _{0}n^{1} \]

### Step-by-Step Solution:

1. **Identify the masses involved in the reaction:**
- Mass of deuterium (\( _{1}H^{2} \)) = 2.014741 amu
- Mass of helium-3 (\( _{2}He^{3} \)) = 3.016977 amu
- Mass of neutron (\( _{0}n^{1} \)) = 1.008987 amu

2. **Calculate the total mass of the reactants:**
- Since there are two deuterium nuclei, the total mass of the reactants is:
\[ \text{Total mass of reactants} = 2 \times \text{mass of deuterium} = 2 \times 2.014741 = 4.029482 \text{ amu} \]

3. **Calculate the total mass of the products:**
- The products of the reaction are one helium-3 nucleus and one neutron:
\[ \text{Total mass of products} = \text{mass of helium-3} + \text{mass of neutron} = 3.016977 + 1.008987 = 4.025964 \text{ amu} \]

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:
\[ \Delta m = \text{Total mass of reactants} - \text{Total mass of products} \]
\[ \Delta m = 4.029482 - 4.025964 = 0.003518 \text{ amu} \]

5. **Convert mass defect to energy (Q-value):**
- The Q-value can be calculated using the formula:
\[ Q = \Delta m \times c^2 \]
- To convert the mass defect from amu to MeV, we use the conversion factor \( 1 \text{ amu} \approx 931 \text{ MeV} \):
\[ Q = 0.003518 \text{ amu} \times 931 \text{ MeV/amu} \approx 3.27 \text{ MeV} \]

### Final Answer:
The Q-value of the reaction is approximately **3.27 MeV**.

---