Consider the reaction `_1^2H+_1^2H=_2^4He+Q`. Mass of the deuterium atom`=2.0141u`. Mass of helium atom `=4.0024u`. This is a nuclear………reaction in which the energy Q released is…………MeV.

Explanation

To solve the problem, we need to determine the energy released in the nuclear reaction given by:

\[ _1^2H + _1^2H \rightarrow _2^4He + Q \]

where the mass of deuterium (\(_1^2H\)) is given as \(2.0141 \, u\) and the mass of helium (\(_2^4He\)) is given as \(4.0024 \, u\).

### Step 1: Calculate the initial mass
The initial mass consists of the masses of the two deuterium nuclei:

\[ \text{Initial mass} = 2 \times \text{mass of deuterium} = 2 \times 2.0141 \, u = 4.0282 \, u \]

### Step 2: Calculate the final mass
The final mass is the mass of the helium nucleus:

\[ \text{Final mass} = \text{mass of helium} = 4.0024 \, u \]

### Step 3: Calculate the change in mass
The change in mass (\(\Delta m\)) is given by the difference between the initial and final masses:

\[ \Delta m = \text{Initial mass} - \text{Final mass} = 4.0282 \, u - 4.0024 \, u = 0.0258 \, u \]

### Step 4: Convert the change in mass to energy
To find the energy released (\(Q\)), we use the mass-energy equivalence principle, which states that:

\[ Q = \Delta m \times 931.4 \, \text{MeV/u} \]

Substituting the value of \(\Delta m\):

\[ Q = 0.0258 \, u \times 931.4 \, \text{MeV/u} \approx 24.05 \, \text{MeV} \]

### Conclusion
The energy \(Q\) released in the reaction is approximately \(24.05 \, \text{MeV}\).

Thus, the reaction is a nuclear fusion reaction, and the energy released is approximately \(24 \, \text{MeV}\).

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