A star initially has `10^40` deuterons. It produces energy via the processes `_1^2H+_1^2Hrarr_1^3H+p` and `_1^2H+_1^3Hrarr_2^4He+n`. Where the masses of the nuclei are
`m(`^2H)=2.014` amu, `m(p)=1.007` amu, `m(n)=1.008` amu and `m(`^4He)=4.001` amu. 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

Options

A. (a) `10^6s`
B. (b) `10^8s`
C. (c) `10^12s`
D. (d) `10^16s`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the reactions and determine the total mass defect
The two reactions given are:
1. \( _1^2H + _1^2H \rightarrow _1^3H + p \)
2. \( _1^2H + _1^3H \rightarrow _2^4He + n \)

Combining these reactions, we can express the overall reaction as:
\[ 3 \, _1^2H \rightarrow _2^4He + n + p \]

### Step 2: Calculate the mass of reactants and products
Using the provided masses:
- Mass of \( _1^2H = 2.014 \) amu
- Mass of \( _1^3H = 3.016 \) amu (not directly given, but can be calculated as \( 2.014 + 1.007 \))
- Mass of \( p = 1.007 \) amu
- Mass of \( n = 1.008 \) amu
- Mass of \( _2^4He = 4.001 \) amu

**Mass of reactants:**
\[ \text{Total mass of reactants} = 3 \times 2.014 = 6.042 \, \text{amu} \]

**Mass of products:**
\[ \text{Total mass of products} = 4.001 + 1.008 + 1.007 = 6.016 \, \text{amu} \]

### Step 3: Calculate the mass defect
The mass defect \( \Delta m \) is given by:
\[ \Delta m = \text{Mass of reactants} - \text{Mass of products} \]
\[ \Delta m = 6.042 - 6.016 = 0.026 \, \text{amu} \]

### Step 4: Convert the mass defect to energy
Using the conversion factor \( 1 \, \text{amu} = 931 \, \text{MeV} \):
\[ \Delta E = \Delta m \times c^2 = 0.026 \, \text{amu} \times 931 \, \text{MeV/amu} \]
\[ \Delta E = 24.218 \, \text{MeV} \]

Converting MeV to Joules (1 MeV = \( 1.6 \times 10^{-13} \) Joules):
\[ \Delta E = 24.218 \times 1.6 \times 10^{-13} \, \text{J} \]
\[ \Delta E \approx 3.87 \times 10^{-12} \, \text{J} \]

### Step 5: Calculate total energy released by \( 10^{40} \) deuterons
Since 3 deuterons produce \( 3.87 \times 10^{-12} \) J:
\[ \text{Energy from } 10^{40} \text{ deuterons} = \frac{10^{40}}{3} \times 3.87 \times 10^{-12} \]
\[ \text{Energy} \approx 1.29 \times 10^{28} \, \text{J} \]

### Step 6: Calculate time to exhaust deuterons
Given the average power \( P = 10^{16} \, \text{W} \):
\[ \text{Time} = \frac{\text{Total Energy}}{\text{Power}} \]
\[ \text{Time} = \frac{1.29 \times 10^{28} \, \text{J}}{10^{16} \, \text{W}} \]
\[ \text{Time} \approx 1.29 \times 10^{12} \, \text{s} \]

### Step 7: Convert time into a more understandable unit
To convert seconds to years:
\[ \text{Time in years} = \frac{1.29 \times 10^{12}}{60 \times 60 \times 24 \times 365} \approx 4.1 \times 10^{4} \, \text{years} \]

### Final Answer
The deuteron supply of the star is exhausted in a time of the order of \( 10^{12} \) seconds or approximately \( 4.1 \times 10^{4} \) years.

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