The binding energies per nucleon for deutrium and helium are `1.1 MeV and 7.0` MeV respectively. The energy in joules will be liberated when `10 ^(6)` deuterons take part in the reaction

Options

A. `18.88 xx 10 ^(-3) J`
B. `18.88 xx 10 ^(-5) J`
C. `18.88 xx 10 ^(-7) J`
D. `18.88 10 ^(-10) J`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Binding Energies
The binding energies per nucleon for deuterium and helium are given as:
- Deuterium: \(1.1 \, \text{MeV}\)
- Helium: \(7.0 \, \text{MeV}\)

### Step 2: Calculate the Total Binding Energy for Deuterium
Since a deuterium nucleus consists of 2 nucleons, the total binding energy for one deuterium nucleus is:
\[ \text{Total Binding Energy for Deuterium} = 2 \times 1.1 \, \text{MeV} = 2.2 \, \text{MeV} \]

### Step 3: Calculate the Total Binding Energy for Helium
A helium nucleus consists of 4 nucleons, so the total binding energy for one helium nucleus is:
\[ \text{Total Binding Energy for Helium} = 4 \times 7.0 \, \text{MeV} = 28.0 \, \text{MeV} \]

### Step 4: Calculate the Mass Defect
The mass defect (\(\Delta m\)) can be calculated as the difference between the total binding energies:
\[ \Delta m = \text{Total Binding Energy for Helium} - \text{Total Binding Energy for Deuterium} \]
\[ \Delta m = 28.0 \, \text{MeV} - 2.2 \, \text{MeV} = 25.8 \, \text{MeV} \]

### Step 5: Calculate the Energy Released for \(10^6\) Deuterons
Since \(10^6\) deuterons are participating in the reaction, the total energy released (\(E\)) can be calculated as:
\[ E = \Delta m \times \frac{10^6}{2} \, \text{MeV} \]
\[ E = 25.8 \, \text{MeV} \times \frac{10^6}{2} = 12.9 \times 10^6 \, \text{MeV} \]

### Step 6: Convert MeV to Joules
To convert MeV to Joules, we use the conversion factor \(1 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J}\):
\[ E = 12.9 \times 10^6 \, \text{MeV} \times 1.6 \times 10^{-13} \, \text{J/MeV} \]
\[ E = 20.64 \times 10^{-7} \, \text{J} = 2.064 \times 10^{-6} \, \text{J} \]

### Final Answer
The energy liberated when \(10^6\) deuterons take part in the reaction is approximately:
\[ \boxed{2.064 \times 10^{-6} \, \text{J}} \]