Calculate the energy generated in kWh, when 100g of `._(3)Li(7)`. are converted into `._(2)He(4)` by proton bombardment. Given mass of `._(3)Li(7)=7.0183a.m.u ,` mass of `._(2)He(4)=4.0040a.m.u`, mass of `._(1)H(1)atom=1.0081a.m.u.`

Explanation

To calculate the energy generated when 100g of Lithium-7 (\(_{3}^{7}\text{Li}\)) is converted into Helium-4 (\(_{2}^{4}\text{He}\)) by proton bombardment, we can follow these steps:

### Step 1: Write the Nuclear Reaction Equation
The nuclear reaction can be represented as:
\[ \text{Li-7} + \text{H-1} \rightarrow \text{He-4} + \text{Energy} \]

### Step 2: Calculate the Mass Defect (\(\Delta m\))
The mass defect is calculated using the formula:
\[ \Delta m = m_{\text{Li}} + m_{\text{H}} - m_{\text{He}} \]
Substituting the given values:
- Mass of \(_{3}^{7}\text{Li} = 7.0183 \, \text{a.m.u}\)
- Mass of \(_{1}^{1}\text{H} = 1.0081 \, \text{a.m.u}\)
- Mass of \(_{2}^{4}\text{He} = 4.0040 \, \text{a.m.u}\)

Calculating:
\[ \Delta m = 7.0183 + 1.0081 - 4.0040 = 4.0224 \, \text{a.m.u} \]

### Step 3: Convert Mass Defect to Energy
The energy released (\(Q\)) can be calculated using the relation:
\[ Q = \Delta m \times 931 \, \text{MeV} \]
For one reaction:
\[ Q = 0.0184 \, \text{a.m.u} \times 931 \, \text{MeV} = 17.13 \, \text{MeV} \]

### Step 4: Calculate the Number of Atoms in 100g of Lithium-7
To find the number of atoms (\(N\)):
\[ N = \frac{\text{mass}}{\text{atomic mass}} \times N_A \]
Where \(N_A = 6.022 \times 10^{23} \, \text{atoms/mol}\) is Avogadro's number.
\[ N = \frac{100 \, \text{g}}{7 \, \text{g/mol}} \times 6.022 \times 10^{23} \approx 8.6 \times 10^{24} \, \text{atoms} \]

### Step 5: Calculate Total Energy Released
The total energy released for 100g of Lithium-7 is:
\[ Q_{\text{total}} = N \times Q \]
Substituting the values:
\[ Q_{\text{total}} = 8.6 \times 10^{24} \times 17.13 \, \text{MeV} \]
Converting MeV to Joules (1 MeV = \(1.6 \times 10^{-13}\) J):
\[ Q_{\text{total}} = 8.6 \times 10^{24} \times 17.13 \times 1.6 \times 10^{-13} \, \text{J} \]

### Step 6: Convert Energy to kWh
To convert Joules to kilowatt-hours:
\[ \text{Energy (kWh)} = \frac{Q_{\text{total}}}{3.6 \times 10^6} \]

### Final Calculation
Putting it all together, we find:
\[ Q_{\text{total}} \approx 6.5475 \times 10^6 \, \text{kWh} \]

### Summary
The energy generated when 100g of Lithium-7 is converted into Helium-4 by proton bombardment is approximately \(6.5475 \times 10^6 \, \text{kWh}\).