`U^(235) + n^(1)rarr` fission product + neutron + `3.2``xx 10^(-11)J`. The energy released , when 1g of `U^(235)` finally undergoes fission , is

Options

A. (a) `12.75xx10^(8)kj`
B. (b) `18.60xx10^(9)kj`
C. (c) `8.21xx10^(7)kj`
D. (d) `6.55xx10^(6)kj`

Explanation

To solve the problem of calculating the energy released when 1 gram of Uranium-235 undergoes fission, we can follow these steps:

### Step 1: Understand the reaction
The fission of Uranium-235 (U-235) when it absorbs a neutron (n) releases energy. The energy released per fission event is given as \(3.2 \times 10^{-11} \, \text{J}\).

### Step 2: Calculate the number of atoms in 1 gram of U-235
The molar mass of U-235 is approximately 235 g/mol. Using Avogadro's number (\(6.022 \times 10^{23}\) atoms/mol), we can find the number of atoms in 1 gram of U-235.

\[ \text{Number of atoms in 1 g of U-235} = \frac{6.022 \times 10^{23} \, \text{atoms/mol}}{235 \, \text{g/mol}} \approx 2.56 \times 10^{21} \, \text{atoms} \]

### Step 3: Calculate the total energy released for 1 gram of U-235
Now, we can find the total energy released when 1 gram of U-235 undergoes fission by multiplying the energy released per fission event by the number of atoms in 1 gram.

\[ \text{Total energy released} = \text{Number of atoms} \times \text{Energy per fission} \]
\[ \text{Total energy released} = 2.56 \times 10^{21} \, \text{atoms} \times 3.2 \times 10^{-11} \, \text{J/atom} \]
\[ \text{Total energy released} \approx 8.19 \times 10^{10} \, \text{J} \]

### Step 4: Convert the energy from Joules to kilojoules
To convert the energy from joules to kilojoules, we divide by 1000.

\[ \text{Total energy released in kJ} = \frac{8.19 \times 10^{10} \, \text{J}}{1000} \approx 8.19 \times 10^{7} \, \text{kJ} \]

### Step 5: Conclusion
The energy released when 1 gram of U-235 undergoes fission is approximately \(8.21 \times 10^{7} \, \text{kJ}\), which corresponds to option C.

### Final Answer
The energy released when 1 gram of U-235 undergoes fission is \(8.21 \times 10^{7} \, \text{kJ}\).

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