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In a nuclear power plant, the process of converting nuclear energy into electrical energy is highly important. This conversion process is often measured by efficiency. Efficiency represents how much of the total energy produced is successfully converted into usable electrical energy. For instance, in a 500 MW nuclear power plant with a 40% efficiency, only 40% of the produced energy is turned into electrical power. To find the total energy generated, divide the electrical power by the efficiency: ewline ewline \( P_{total} = \frac{P_{electrical}}{\text{Efficiency}} = \frac{500 \text{ MW}}{0.40} = 1250 \text{ MW} \). This calculation shows us that for 500 MW of electrical power, the plant produces 1250 MW total energy.

Uranium-235 is a key element used in nuclear power plants because of its ability to undergo fission. Fission is a nuclear reaction where the nucleus of an atom splits into two or more smaller nuclei, along with a few neutrons and a large amount of energy. This energy is what powers the turbines that generate electricity. When a Uranium-235 atom absorbs a neutron, it becomes unstable and splits. The energy released per fission of Uranium-235 is equivalent to about 200 MeV (Mega Electron Volts). To understand this better, we can convert these units to more familiar terms like Joules. ewline ewline Since 1 eV is equal to \(1.60218 \times 10^{-19} \text{ Joules} \), we can calculate: ewline ewline 200 MeV = 200 \times 10^6 eV \times 1.60218 \times 10^{-19} J/eV = 3.20436 \times 10^{-11} J.

Each fission reaction of Uranium-235 releases a tremendous amount of energy. Specifically, it releases \(3.20436 \times 10^{-11} \text{ Joules} \). This might sound small, but consider that in a nuclear power plant, this happens countless times every second. To supply a continuous output of 500 MW (with a plant operating at 40% efficiency), the rate of these reactions is staggering. First, we convert the electric power to total energy considering efficiency. Then, we find the required energy released per second in Joules by converting 1250 MW to 1.25 \times 10^9 J/s. Next, we look at how many of these tiny reactions are required to meet this energy demand. ewline ewline This is done using: ewline ewline \( \text{Number of fissions per second} = \frac{1.25 \times 10^9 \text{ J/s}}{3.20436 \times 10^{-11} \text{ J}} \approx 3.90 \times 10^{19} \text{ fissions/s} \).

Calculating the power output and the required material in a nuclear plant is crucial. For example, to determine how much Uranium-235 is needed per second, we have our conversion requirements. We already know the number of fissions required per second, approximately \(3.90 \times 10^{19} \text{ fissions/s} \). ewline Now, we need to figure out how much mass this involves. Moles are a useful unit here: one mole of Uranium-235 weighs 235 grams and contains Avogadro's number of atoms (6.022 \times 10^{23} atoms/mole). Calculating moles per second, we get: ewline ewline \( \frac{3.90 \times 10^{19}}{6.022 \times 10^{23}} \approx 6.48 \times 10^{-5} \text{ moles/s} \). Finally, multiply this by the atomic mass to find the mass of Uranium-235 required every second: ewline ewline \( \text{Mass of Uranium-235} = 6.48 \times 10^{-5} \times 235 \approx 0.0152 \text{ grams/s} \). Thus, a very small amount of Uranium-235 (about 0.0152 grams) must undergo fission each second to keep the plant running.