The energy released during the fission of 1kg of `U^(235)` is a `E_1` and that product during the fusion of 1 kg of hydrogen is `E_2` . If energy released per fission of Uranium - 235 is 200 MeV and that per fusion of hydrogen is 24.7 MeV , then the ratio `E_2/E_1` is

Options

A. 2
B. 7
C. 10
D. 20

Answer

To solve the problem, we need to calculate the energy released during the fission of 1 kg of Uranium-235 (denoted as \( E_1 \)) and the energy released during the fusion of 1 kg of hydrogen (denoted as \( E_2 \)). We will then find the ratio \( \frac{E_2}{E_1} \).

### Step 1: Calculate the number of fissions in 1 kg of Uranium-235

1. **Mass of Uranium-235**: 1 kg = 1000 g
2. **Atomic mass of Uranium-235**: Approximately 235 g/mol
3. **Number of moles of Uranium-235 in 1 kg**:
\[ \text{Number of moles} = \frac{1000 \text{ g}}{235 \text{ g/mol}} \approx 4.26 \text{ moles} \]
4. **Number of atoms in 1 kg of Uranium-235** (using Avogadro's number \( N_A \approx 6.022 \times 10^{23} \text{ atoms/mol} \)):
\[ \text{Number of atoms} = 4.26 \text{ moles} \times 6.022 \times 10^{23} \text{ atoms/mol} \approx 2.56 \times 10^{24} \text{ atoms} \]
5. **Number of fissions**: Each fission event corresponds to one atom of Uranium-235 undergoing fission, so the number of fissions \( N_f \) is approximately \( 2.56 \times 10^{24} \).

### Step 2: Calculate the energy released during fission (\( E_1 \))

1. **Energy released per fission of Uranium-235**: 200 MeV
2. **Total energy released during fission of 1 kg of Uranium-235**:
\[ E_1 = N_f \times \text{Energy per fission} = 2.56 \times 10^{24} \times 200 \text{ MeV} \]
\[ E_1 = 5.12 \times 10^{26} \text{ MeV} \]

### Step 3: Calculate the number of fusions in 1 kg of hydrogen

1. **Mass of hydrogen required for one fusion**: 4 g (since two hydrogen nuclei combine to form one helium nucleus).
2. **Number of fusions \( N_fusion \)** in 1 kg of hydrogen:
\[ N_fusion = \frac{1000 \text{ g}}{4 \text{ g}} = 250 \]

### Step 4: Calculate the energy released during fusion (\( E_2 \))

1. **Energy released per fusion of hydrogen**: 24.7 MeV
2. **Total energy released during fusion of 1 kg of hydrogen**:
\[ E_2 = N_fusion \times \text{Energy per fusion} = 250 \times 24.7 \text{ MeV} \]
\[ E_2 = 6175 \text{ MeV} \]

### Step 5: Calculate the ratio \( \frac{E_2}{E_1} \)

1. **Using the values calculated for \( E_1 \) and \( E_2 \)**:
\[ \frac{E_2}{E_1} = \frac{6175 \text{ MeV}}{5.12 \times 10^{26} \text{ MeV}} \approx 7 \]

### Final Answer

The ratio \( \frac{E_2}{E_1} \) is approximately **7**.

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