A star initially has `10^40` deuterons. It produces energy via the processes `_1H^2+_1H^2rarr_1H^3+p` and `_1H^2+_1H^3rarr_2He^4+n`. If the average power radiated by the star is `10^16` W, the deuteron supply of the star is exhausted in a time of the order of
(a) `10^6s` (b) `10^8s` (c) `10^12s`
The masses of the nuclei are as follows
`M(H^2)=2.014` amu, `M(n)=1.008` amu,
`M(p)=1.007` amu,`M(He^4)=4.001`amu

Explanation

To solve the problem step by step, we need to analyze the fusion processes occurring in the star and calculate the time it takes for the deuteron supply to be exhausted.

### Step 1: Understand the Fusion Processes
The star undergoes two fusion reactions:
1. \( _1H^2 + _1H^2 \rightarrow _1H^3 + p \) (Deuteron fusion producing Tritium and a Proton)
2. \( _1H^2 + _1H^3 \rightarrow _2He^4 + n \) (Deuteron and Tritium fusion producing Helium and a Neutron)

### Step 2: Write the Overall Reaction
By combining the two reactions, we can derive an overall reaction:
- From the first reaction, we consume 2 deuterons and produce 1 tritium and 1 proton.
- From the second reaction, we consume 1 deuteron and 1 tritium to produce 1 helium and 1 neutron.

Combining these gives:
\[ 3 \, _1H^2 \rightarrow 2 \, _2He^4 + 1 \, p + 1 \, n \]

### Step 3: Calculate the Mass Defect
Now, we calculate the mass defect \( \Delta M \):
- Mass of reactants: \( 3 \times M(H^2) = 3 \times 2.014 \, \text{amu} = 6.042 \, \text{amu} \)
- Mass of products:
- Mass of Helium: \( 4.001 \, \text{amu} \)
- Mass of Proton: \( 1.007 \, \text{amu} \)
- Mass of Neutron: \( 1.008 \, \text{amu} \)

Thus, the total mass of products is:
\[ M_{products} = 4.001 + 1.007 + 1.008 = 6.016 \, \text{amu} \]

Now, calculate the mass defect:
\[ \Delta M = M_{reactants} - M_{products} = 6.042 - 6.016 = 0.026 \, \text{amu} \]

### Step 4: Convert Mass Defect to Energy
Using the conversion factor \( 1 \, \text{amu} \approx 931 \, \text{MeV/c}^2 \):
\[ E = \Delta M \times 931 \, \text{MeV} = 0.026 \times 931 \approx 24.206 \, \text{MeV} \]

### Step 5: Calculate Total Energy Released
Since 3 deuterons produce this energy, the energy released per deuteron is:
\[ E_{per \, deuteron} = \frac{24.206}{3} \approx 8.068 \, \text{MeV} \]

For \( 10^{40} \) deuterons, the total energy released is:
\[ E_{total} = 8.068 \times 10^{40} \, \text{MeV} \]

### Step 6: Convert Energy to Joules
To convert MeV to Joules, use the conversion \( 1 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J} \):
\[ E_{total} = 8.068 \times 10^{40} \times 1.6 \times 10^{-13} \approx 1.29 \times 10^{28} \, \text{J} \]

### Step 7: Calculate Time for Deuteron Exhaustion
Power is defined as energy per unit time:
\[ P = \frac{E}{T} \]
Thus, the time \( T \) can be calculated as:
\[ T = \frac{E}{P} = \frac{1.29 \times 10^{28} \, \text{J}}{10^{16} \, \text{W}} = 1.29 \times 10^{12} \, \text{s} \]

### Conclusion
The time required to exhaust the deuteron supply is of the order of \( 10^{12} \, \text{s} \).

### Final Answer
(c) \( 10^{12} \, \text{s} \)

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