In the nuclear raction `._1H^2 +._1H^2 rarr ._2He^3 +._0n^1` if the mass of the deuterium atom `=2.014741 am u`, mass of `._2He^3` atom `=3.016977 am u`, and mass of neutron `=1.008987 am u`, 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 find the Q value of the nuclear reaction \( _1H^2 + _1H^2 \rightarrow _2He^3 + _0n^1 \), we will follow these steps:

### Step 1: Identify the masses of the reactants and products
- The mass of deuterium (\( _1H^2 \)) is given as \( 2.014741 \, \text{amu} \).
- The mass of helium-3 (\( _2He^3 \)) is given as \( 3.016977 \, \text{amu} \).
- The mass of a neutron (\( _0n^1 \)) is given as \( 1.008987 \, \text{amu} \).

### Step 2: Calculate the total mass of the reactants
The total mass of the reactants (two deuterium nuclei) is:
\[ \text{Mass of reactants} = 2 \times \text{mass of } _1H^2 = 2 \times 2.014741 \, \text{amu} = 4.029482 \, \text{amu} \]

### Step 3: Calculate the total mass of the products
The total mass of the products (one helium-3 nucleus and one neutron) is:
\[ \text{Mass of products} = \text{mass of } _2He^3 + \text{mass of } _0n^1 = 3.016977 \, \text{amu} + 1.008987 \, \text{amu} = 4.025964 \, \text{amu} \]

### 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:
\[ \Delta m = \text{Mass of reactants} - \text{Mass of products} = 4.029482 \, \text{amu} - 4.025964 \, \text{amu} = 0.003518 \, \text{amu} \]

### Step 5: Convert the mass defect to energy (Q value)
Using the conversion factor \( 1 \, \text{amu} = 931.5 \, \text{MeV} \), we can find the Q value:
\[ Q = \Delta m \times 931.5 \, \text{MeV} = 0.003518 \, \text{amu} \times 931.5 \, \text{MeV/amu} \approx 3.27 \, \text{MeV} \]

### Conclusion
The Q value of the reaction is approximately \( 3.27 \, \text{MeV} \).

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