Assuming that about `200 MeV` of energy is released per fission of `._(92)U^(235)` nuceli, the mass of `U^(235)` consumed per day in a fission ractor of power `1` megawatt will be approximately .

Options

A. `10^(-2)g`
B. `1 g`
C. `100 g`
D. `10,000 g`

Explanation

To find the mass of \( U^{235} \) consumed per day in a fission reactor of power \( 1 \) megawatt, we can follow these steps:

### Step 1: Calculate the total energy produced in one day
The power of the reactor is given as \( 1 \) megawatt, which is equivalent to \( 10^6 \) watts. The time in one day can be calculated as:
\[ \text{Time} = 24 \text{ hours} \times 3600 \text{ seconds/hour} = 86400 \text{ seconds} \]
Now, we can calculate the total energy produced in one day:
\[ \text{Energy} = \text{Power} \times \text{Time} = 10^6 \text{ W} \times 86400 \text{ s} = 8.64 \times 10^{10} \text{ J} \]

### Step 2: Convert energy released per fission from MeV to Joules
The energy released per fission of \( U^{235} \) is given as \( 200 \) MeV. To convert this to joules, we use the conversion factor \( 1 \text{ MeV} = 1.6 \times 10^{-13} \text{ J} \):
\[ \text{Energy per fission} = 200 \text{ MeV} \times 1.6 \times 10^{-13} \text{ J/MeV} = 3.2 \times 10^{-11} \text{ J} \]

### Step 3: Calculate the number of fissions occurring in one day
To find the number of fissions that occur in one day, we divide the total energy produced by the energy released per fission:
\[ \text{Number of fissions} = \frac{\text{Total Energy}}{\text{Energy per fission}} = \frac{8.64 \times 10^{10} \text{ J}}{3.2 \times 10^{-11} \text{ J}} = 2.7 \times 10^{21} \]

### Step 4: Calculate the mass of \( U^{235} \) consumed
To find the mass consumed, we need to use Avogadro's number \( N_A \) and the molar mass of \( U^{235} \). The molar mass of \( U^{235} \) is approximately \( 235 \text{ g/mol} \) and Avogadro's number is \( 6.023 \times 10^{23} \text{ atoms/mol} \).

First, we find the number of moles of \( U^{235} \) consumed:
\[ \text{Number of moles} = \frac{\text{Number of fissions}}{N_A} = \frac{2.7 \times 10^{21}}{6.023 \times 10^{23}} \approx 4.48 \times 10^{-3} \text{ moles} \]

Now, we can calculate the mass consumed:
\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} = 4.48 \times 10^{-3} \text{ moles} \times 235 \text{ g/mol} \approx 1.05 \text{ g} \]

### Final Answer
The mass of \( U^{235} \) consumed per day in a fission reactor of power \( 1 \) megawatt is approximately \( 1 \text{ g} \).

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