c2v2zw4k
1 year ago
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Uranium-235: 200 MeV per fission |
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Uranium-233: 197 MeV per fission |
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Plutonium-239: 210 MeV per fission |
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The amount of energy needed to reach 15,111 megatons is: |
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E = 15.111 × 10^6 × 4.184 × 10^12 Joules, where 4.184 × 10^12 Joules |
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is the conversion of 1 megaton to Joules. |
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Now we can calculate the number of fissions needed to |
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each element, considering the efficiency of 0.147: |
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For Uranium-235: |
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N_fissions_Uranium-235 = E / (Efficiency × Energy per fission) |
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N_fissions_Uranium-235 = 15.111 × 10^6 × 4.184 × 10^12 / (0.147 × 200) |
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N_fissions_Uranium-235 ≈ 5,351 × 10^17 fissions |
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For Uranium-233: |
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N_fissions_Uranium-233 = E / (Efficiency × Energy per fission) |
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N_fissions_Uranium-233 = 15.111 × 10^6 × 4.184 × 10^12 / (0.147 × 197) |
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N_fissions_Uranium-233 ≈ 5,442 × 10^17 fissions |
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For Plutonium-239: |
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N_fissions_Plutonium-239 = E / (Efficiency × Energy per fission) |
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N_fissions_Plutonium-239 = 15.111 × 10^6 × 4.184 × 10^12 / (0.147 × 210) |
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N_fissions_Plutonium-239 ≈ 5057 × 10^17 fissions |
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Finally, we can calculate the amount of mass of each element needed, |
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considering the number of fissions and the molar mass of each element: |
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For Uranium-235: |
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m_Uranium-235 = N_fissions_Uranium-235 × Molar mass of Uranium-235 / Avogadro's number |
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The molar mass of Uranium-235 is approximately 235 g/mol. |
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For Uranium-233: |
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m_Uranium-233 = N_fissions_Uranium-233 × Molar mass of Uranium-233 / Avogadro's number |
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The molar mass of Uranium-233 is approximately 233 g/mol. |
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For Plutonium-239: |
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m_Plutonium-239 = N_fissions_Plutonium-239 × Molar Mass of Plutonium-239 / Avogadro's Number |
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The molar mass of Plutonium-239 is approximately 239 g/mol. |
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Now we can calculate the masses: |
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For Uranium-235: |
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m_Uranium-235 = 5.351 × 10^17 × 235 / 6.022 × 10^23 |
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m_Uranium-235 ≈ 21.04 kg |
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For Uranium-233: |
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m_Uranium-233 = 5.442 × 10^17 × 233 / 6.022 × 10^23 |
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m_Uranium-233 ≈ 21.43 kg |
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For Plutonium-239: |
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m_Plutonium-239 = 5.057 × 10^17 × 239 / 6.022 × 10^23 |
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m_Plutonium-239 ≈ 19.97 |
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Now we can calculate the required masses of each element |
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to reach 15,111 megatons of energy: |
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For Uranium-235: |
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m_Uranium-235 = 5.351 × 10^17 × 235 / 6.022 × 10^23 |
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m_Uranium-235 ≈ 21.04 kg |
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For Uranium-233: |
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m_Uranium-233 = 5.442 × 10^17 × 233 / 6.022 × 10^23 |
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m_Uranium-233 ≈ 21.43 kg |
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For Plutonium-239: |
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m_Plutonium-239 = 5.057 × 10^17 × 239 / 6.022 × 10^23 |
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m_Plutonium-239 ≈ 19.97 kg |
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Therefore, to reach 15,111 megatons of energy, |
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It would take approximately: |
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21.04 kg of Uranium-235 |
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21.43 kg of Uranium-233 |
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19.97 kg of Plutonium-239 |
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