To solve the problem, we need to calculate how long a 100 W lamp can glow using the energy released from the fusion reaction involving 2 kg of \( _1^2A \).
### Step-by-Step Solution:
1. **Identify the Reaction**:
The fusion reaction is given as:
\[
2 \, _1^2A \rightarrow _2^3B + n + 3.27 \, \text{MeV}
\]
This means that 2 atoms of \( _1^2A \) produce 1 atom of \( _2^3B \), 1 neutron, and release 3.27 MeV of energy.
2. **Convert Mass of \( _1^2A \) to Number of Atoms**:
The molar mass of \( _1^2A \) is 2 g/mol. Therefore, the number of moles in 2 kg (2000 g) of \( _1^2A \) is:
\[
\text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{2000 \, \text{g}}{2 \, \text{g/mol}} = 1000 \, \text{mol}
\]
Using Avogadro's number (\( 6.022 \times 10^{23} \) atoms/mol), the total number of atoms is:
\[
\text{Number of atoms} = 1000 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} \approx 6.022 \times 10^{26} \, \text{atoms}
\]
3. **Calculate the Number of Reactions**:
Since 2 atoms of \( _1^2A \) are needed for one reaction, the number of reactions that can occur is:
\[
\text{Number of reactions} = \frac{6.022 \times 10^{26}}{2} \approx 3.011 \times 10^{26}
\]
4. **Calculate Total Energy Released**:
Each reaction releases 3.27 MeV. The total energy released in joules is calculated as follows:
\[
\text{Total energy (in MeV)} = 3.011 \times 10^{26} \times 3.27 \, \text{MeV} \approx 9.85 \times 10^{26} \, \text{MeV}
\]
To convert MeV to joules, we use the conversion factor \( 1 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J} \):
\[
\text{Total energy (in J)} = 9.85 \times 10^{26} \times 1.6 \times 10^{-13} \approx 1.576 \times 10^{14} \, \text{J}
\]
5. **Calculate Time for the Lamp to Glow**:
The power of the lamp is given as 100 W (which is 100 J/s). The time \( t \) the lamp can glow is given by:
\[
t = \frac{\text{Total energy}}{\text{Power}} = \frac{1.576 \times 10^{14} \, \text{J}}{100 \, \text{W}} = 1.576 \times 10^{12} \, \text{s}
\]
6. **Convert Time from Seconds to Years**:
To convert seconds to years:
\[
t \text{ (in years)} = \frac{1.576 \times 10^{12} \, \text{s}}{365 \times 24 \times 3600} \approx 50.0 \times 10^{4} \, \text{years} \approx 5 \times 10^{4} \, \text{years}
\]
### Final Answer:
The lamp will glow for approximately \( 5 \times 10^{4} \) years.
