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Step-by-Step Solution
Step 1: Write down the known data
Power of the reactor, $P = 10^9 \text{ watt}$ (i.e., $10^9 \text{ J/s}$)
Velocity of light, $c = 3 \times 10^8 \text{ m/s}$
Time for which we want the mass consumption, $\Delta t = 1 \text{ hour} = 3600 \text{ s}$
Step 2: Write the relation connecting mass and energy
According to Einstein’s mass-energy equivalence,
$$
E = \Delta m \, c^2.
$$
When a reactor operates at power $P$, the energy produced per unit time is given by
$$
P = \frac{E}{\Delta t} = \frac{\Delta m \, c^2}{\Delta t}.
$$
Rearranging for $\Delta m$:
$$
\Delta m = \frac{P \,\Delta t}{c^2}.
$$
Step 3: Substitute the numerical values
Substitute $P = 10^9 \, \text{J/s}$, $\Delta t = 3600 \, \text{s}$, and $c = 3 \times 10^8 \, \text{m/s}$ into the formula:
$$
\Delta m = \frac{(10^9)\times(3600)}{(3 \times 10^8)^2}.
$$
First, calculate $(3 \times 10^8)^2$:
$$
(3 \times 10^8)^2 = 9 \times 10^{16}.
$$
Next, multiply $10^9 \times 3600$:
$$
10^9 \times 3600 = 3.6 \times 10^{12}.
$$
Therefore,
$$
\Delta m = \frac{3.6 \times 10^{12}}{9 \times 10^{16}} \,\text{kg}.
$$
Step 4: Simplify the expression
Divide the numbers and powers of 10 carefully:
$$
\Delta m = \frac{3.6}{9} \times 10^{12 - 16} = 0.4 \times 10^{-4} \,\text{kg}.
$$
Which is
$$
\Delta m = 4 \times 10^{-5} \,\text{kg}.
$$
Step 5: Convert the mass into grams
Since $1 \,\text{kg} = 1000 \,\text{g}$,
$$
4 \times 10^{-5} \,\text{kg} = 4 \times 10^{-5} \times 1000 \,\text{g} = 4 \times 10^{-2} \,\text{g}.
$$
Hence, the mass consumed in one hour is $4 \times 10^{-2} \,\text{g}.$
Answer
The mass of the fuel consumed per hour in the reactor is
$$
\boxed{4 \times 10^{-2} \,\text{g}.}
$$
