The following fusion reaction take place `2_1^2Ararr_2^3B+n+3.27` MeV. If 2 kg of `._1^2A` is subjected to the above reaction, the energy released is used to light a 100 W light a lamp, how long will the lamp glow ?

Options

A. `7xx10^3` years
B. `3xx10^5` years
C. `5xx10^4` years
D. `2xx10^6` years

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Calculate the number of moles of \( ^2_1A \)
Given that the mass of \( ^2_1A \) is 2 kg, we first convert this to grams:
\[ \text{Mass} = 2 \, \text{kg} = 2000 \, \text{g} \]
The molecular mass of \( ^2_1A \) is 2 g/mol.

Now, we can calculate the number of moles (\( n \)):
\[ n = \frac{\text{mass}}{\text{molecular mass}} = \frac{2000 \, \text{g}}{2 \, \text{g/mol}} = 1000 \, \text{mol} \]

### Step 2: Calculate the number of atoms
Using Avogadro's number (\( N_A = 6.022 \times 10^{23} \, \text{atoms/mol} \)), we can find the total number of atoms:
\[ \text{Number of atoms} = n \times N_A = 1000 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} = 6.022 \times 10^{26} \, \text{atoms} \]

### Step 3: Calculate the total energy released
From the reaction, we know that 2 atoms of \( ^2_1A \) release 3.27 MeV of energy. Therefore, the energy released per atom is:
\[ \text{Energy per atom} = \frac{3.27 \, \text{MeV}}{2} = 1.635 \, \text{MeV} \]
To convert MeV to joules, we use the conversion factor \( 1 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J} \):
\[ \text{Energy per atom in joules} = 1.635 \, \text{MeV} \times 1.6 \times 10^{-13} \, \text{J/MeV} = 2.616 \times 10^{-13} \, \text{J} \]
Now, the total energy released by all atoms is:
\[ \text{Total energy} = \text{Number of atoms} \times \text{Energy per atom} = 6.022 \times 10^{26} \times 2.616 \times 10^{-13} \, \text{J} \approx 1.577 \times 10^{14} \, \text{J} \]

### Step 4: Calculate the time the lamp will glow
The power of the lamp is given as 100 W. The time (\( t \)) for which the lamp will glow can be calculated using the formula:
\[ t = \frac{\text{Total energy}}{\text{Power}} = \frac{1.577 \times 10^{14} \, \text{J}}{100 \, \text{W}} = 1.577 \times 10^{12} \, \text{s} \]

### Step 5: Convert seconds to years
To convert seconds to years, we use the conversion factor \( 1 \, \text{year} \approx 3.154 \times 10^7 \, \text{s} \):
\[ t \, (\text{in years}) = \frac{1.577 \times 10^{12} \, \text{s}}{3.154 \times 10^7 \, \text{s/year}} \approx 5.00 \times 10^4 \, \text{years} \]

### Final Answer
The lamp will glow for approximately \( 5.00 \times 10^4 \) years.

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