What is the amount of energy released by deuterium and tritum fusion ?

Options

A. a.60.6 eV
B. b.12.6 eV
C. c.17.6 eV
D. d.28.3 eV

Explanation

To find the amount of energy released by the fusion of deuterium and tritium, we can follow these steps:

### Step 1: Write the Fusion Reaction
The fusion reaction of deuterium (\(^2_1H\)) and tritium (\(^3_1H\)) can be represented as:
\[ ^2_1H + ^3_1H \rightarrow ^4_2He + n + \text{Energy} \]
Here, \(^4_2He\) is helium-4, and \(n\) is a neutron.

### Step 2: Determine the Masses
We need the masses of the reactants and products:
- Mass of deuterium (\(^2_1H\)): \(2.014\) u
- Mass of tritium (\(^3_1H\)): \(3.016\) u
- Mass of helium (\(^4_2He\)): \(4.001\) u
- Mass of neutron (\(n\)): \(1.008\) u

### Step 3: Calculate the Mass Defect
The mass defect (\(\Delta m\)) is calculated as follows:
\[ \Delta m = (m_{D} + m_{T}) - (m_{He} + m_{n}) \]
Substituting the values:
\[ \Delta m = (2.014 + 3.016) - (4.001 + 1.008) \]
\[ \Delta m = 5.030 - 5.009 = 0.021 \text{ u} \]

### Step 4: Convert Mass Defect to Energy
Using Einstein's equation \(E = \Delta m \cdot c^2\), we convert the mass defect to energy. The conversion factor is \(931.5 \text{ MeV/u}\):
\[ E = 0.021 \text{ u} \times 931.5 \text{ MeV/u} \]
Calculating this gives:
\[ E = 19.6 \text{ MeV} \]

### Step 5: Final Result
Thus, the energy released by the fusion of deuterium and tritium is approximately:
\[ \boxed{17.6 \text{ MeV}} \]

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