Imagine that a reactor converts all given mass into energy and that it operates at a power level of `10^9` watt. The mass of the fuel consumed per hour in the reactor will be (velocity of light, c is `3 xx 10^8` m/s)

Options

A. 0.96 g
B. 0.8 g
C. `4 xx 10^(-2)`g
D. `6.6 xx 10^(-5)` g

Explanation

To find the mass of fuel consumed per hour in a reactor that converts all given mass into energy and operates at a power level of \(10^9\) watts, we can follow these steps:

### Step 1: Understand the relationship between power, energy, and mass
Power (\(P\)) is defined as the rate of energy (\(E\)) transfer per unit time (\(t\)):
\[ P = \frac{E}{t} \]
Given that the reactor operates at a power level of \(10^9\) watts, we can express this as:
\[ E = P \cdot t \]

### Step 2: Relate energy to mass using Einstein's equation
According to Einstein's mass-energy equivalence principle, the energy produced from a mass (\(m\)) can be expressed as:
\[ E = mc^2 \]
where \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)).

### Step 3: Calculate the energy produced in one hour
Since we want to find the mass consumed per hour, we need to calculate the total energy produced in one hour:
\[ t = 1 \, \text{hour} = 3600 \, \text{seconds} \]
Thus, the energy produced in one hour is:
\[ E = P \cdot t = 10^9 \, \text{W} \cdot 3600 \, \text{s} = 3.6 \times 10^{12} \, \text{J} \]

### Step 4: Relate the energy back to mass
Now, we can relate this energy back to mass using the equation \(E = mc^2\):
\[ m = \frac{E}{c^2} \]
Substituting the values we have:
\[ m = \frac{3.6 \times 10^{12} \, \text{J}}{(3 \times 10^8 \, \text{m/s})^2} \]

### Step 5: Calculate \(c^2\)
Calculating \(c^2\):
\[ c^2 = (3 \times 10^8)^2 = 9 \times 10^{16} \, \text{m}^2/\text{s}^2 \]

### Step 6: Substitute \(c^2\) back into the mass equation
Now substituting \(c^2\) into the mass equation:
\[ m = \frac{3.6 \times 10^{12}}{9 \times 10^{16}} = 4 \times 10^{-5} \, \text{kg} \]

### Step 7: Convert mass from kg to grams
Since \(1 \, \text{kg} = 1000 \, \text{grams}\):
\[ m = 4 \times 10^{-5} \, \text{kg} \times 1000 \, \text{g/kg} = 4 \times 10^{-2} \, \text{g} \]

### Final Answer
The mass of the fuel consumed per hour in the reactor is:
\[ \boxed{4 \times 10^{-2} \, \text{grams}} \]