How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? The fusion reaction can be taken as `._1H^2+._1H^2to ._1H^3+n+3.17MeV`

Options

A. `2.4xx10^6` years
B. `7.4xx10^4` years
C. `1.6xx10^6` years
D. `4.9xx10^4` years

Explanation

To solve the problem of how long a 100W electric lamp can be kept glowing by the fusion of 2.0 kg of deuterium, we will follow these steps:

### Step 1: Understand the Fusion Reaction
The fusion reaction given is:
\[ _1H^2 + _1H^2 \rightarrow _1H^3 + n + 3.17 \text{ MeV} \]
This means that two deuterium nuclei fuse to produce a tritium nucleus, a neutron, and release 3.17 MeV of energy.

### Step 2: Calculate the Total Energy Released
First, we need to find out how many fusion reactions can occur with 2.0 kg of deuterium.

1. **Calculate the number of moles of deuterium:**
- The molar mass of deuterium (D) is approximately 2 g/mol.
- Total mass of deuterium = 2000 g.
- Number of moles (n) of deuterium:
\[ n = \frac{\text{Total mass}}{\text{Molar mass}} = \frac{2000 \text{ g}}{2 \text{ g/mol}} = 1000 \text{ mol} \]

2. **Calculate the total number of deuterium atoms:**
- Using Avogadro's number (\(6.022 \times 10^{23}\) atoms/mol):
\[ \text{Total number of deuterium atoms} = 1000 \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol} = 6.022 \times 10^{26} \text{ atoms} \]

3. **Calculate the number of fusion reactions:**
- Each fusion reaction consumes 2 deuterium atoms, so the number of reactions (R) is:
\[ R = \frac{\text{Total number of deuterium atoms}}{2} = \frac{6.022 \times 10^{26}}{2} = 3.011 \times 10^{26} \text{ reactions} \]

4. **Calculate the total energy released:**
- Energy released per reaction = 3.17 MeV.
- Convert MeV to Joules (1 MeV = \(1.6 \times 10^{-13}\) Joules):
\[ \text{Energy per reaction in Joules} = 3.17 \text{ MeV} \times 1.6 \times 10^{-13} \text{ J/MeV} = 5.072 \times 10^{-13} \text{ J} \]
- Total energy (E) released:
\[ E = R \times \text{Energy per reaction} = 3.011 \times 10^{26} \times 5.072 \times 10^{-13} \text{ J} \approx 1.527 \times 10^{14} \text{ J} \]

### Step 3: Calculate the Time the Lamp Can Glow
Using the formula:
\[ \text{Energy} = \text{Power} \times \text{Time} \]
We can rearrange this to find time:
\[ \text{Time} = \frac{\text{Energy}}{\text{Power}} = \frac{1.527 \times 10^{14} \text{ J}}{100 \text{ W}} = 1.527 \times 10^{12} \text{ seconds} \]

### Step 4: Convert Time to Years
To convert seconds into years:
\[ \text{Time in years} = \frac{1.527 \times 10^{12} \text{ seconds}}{3.15 \times 10^{7} \text{ seconds/year}} \approx 4.85 \times 10^{4} \text{ years} \]

### Final Answer
The electric lamp can be kept glowing for approximately \(4.85 \times 10^{4}\) years.

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