You can not select more than 25 topics
Topics must start with a letter or number, can include dashes ('-') and can be up to 35 characters long.
80 lines
2.5 KiB
80 lines
2.5 KiB
1 year ago
|
Uranium-235: 200 MeV per fission
|
||
|
Uranium-233: 197 MeV per fission
|
||
|
Plutonium-239: 210 MeV per fission
|
||
|
|
||
|
The amount of energy needed to reach 15,111 megatons is:
|
||
|
E = 15.111 × 10^6 × 4.184 × 10^12 Joules, where 4.184 × 10^12 Joules
|
||
|
is the conversion of 1 megaton to Joules.
|
||
|
|
||
|
Now we can calculate the number of fissions needed to
|
||
|
each element, considering the efficiency of 0.147:
|
||
|
|
||
|
For Uranium-235:
|
||
|
N_fissions_Uranium-235 = E / (Efficiency × Energy per fission)
|
||
|
N_fissions_Uranium-235 = 15.111 × 10^6 × 4.184 × 10^12 / (0.147 × 200)
|
||
|
N_fissions_Uranium-235 ≈ 5,351 × 10^17 fissions
|
||
|
|
||
|
For Uranium-233:
|
||
|
N_fissions_Uranium-233 = E / (Efficiency × Energy per fission)
|
||
|
N_fissions_Uranium-233 = 15.111 × 10^6 × 4.184 × 10^12 / (0.147 × 197)
|
||
|
N_fissions_Uranium-233 ≈ 5,442 × 10^17 fissions
|
||
|
|
||
|
For Plutonium-239:
|
||
|
N_fissions_Plutonium-239 = E / (Efficiency × Energy per fission)
|
||
|
N_fissions_Plutonium-239 = 15.111 × 10^6 × 4.184 × 10^12 / (0.147 × 210)
|
||
|
N_fissions_Plutonium-239 ≈ 5057 × 10^17 fissions
|
||
|
|
||
|
Finally, we can calculate the amount of mass of each element needed,
|
||
|
considering the number of fissions and the molar mass of each element:
|
||
|
|
||
|
For Uranium-235:
|
||
|
m_Uranium-235 = N_fissions_Uranium-235 × Molar mass of Uranium-235 / Avogadro's number
|
||
|
|
||
|
The molar mass of Uranium-235 is approximately 235 g/mol.
|
||
|
|
||
|
For Uranium-233:
|
||
|
m_Uranium-233 = N_fissions_Uranium-233 × Molar mass of Uranium-233 / Avogadro's number
|
||
|
|
||
|
The molar mass of Uranium-233 is approximately 233 g/mol.
|
||
|
|
||
|
For Plutonium-239:
|
||
|
m_Plutonium-239 = N_fissions_Plutonium-239 × Molar Mass of Plutonium-239 / Avogadro's Number
|
||
|
|
||
|
The molar mass of Plutonium-239 is approximately 239 g/mol.
|
||
|
|
||
|
Now we can calculate the masses:
|
||
|
|
||
|
For Uranium-235:
|
||
|
m_Uranium-235 = 5.351 × 10^17 × 235 / 6.022 × 10^23
|
||
|
m_Uranium-235 ≈ 21.04 kg
|
||
|
|
||
|
For Uranium-233:
|
||
|
m_Uranium-233 = 5.442 × 10^17 × 233 / 6.022 × 10^23
|
||
|
m_Uranium-233 ≈ 21.43 kg
|
||
|
|
||
|
For Plutonium-239:
|
||
|
m_Plutonium-239 = 5.057 × 10^17 × 239 / 6.022 × 10^23
|
||
|
m_Plutonium-239 ≈ 19.97
|
||
|
|
||
|
Now we can calculate the required masses of each element
|
||
|
to reach 15,111 megatons of energy:
|
||
|
|
||
|
For Uranium-235:
|
||
|
m_Uranium-235 = 5.351 × 10^17 × 235 / 6.022 × 10^23
|
||
|
m_Uranium-235 ≈ 21.04 kg
|
||
|
|
||
|
For Uranium-233:
|
||
|
m_Uranium-233 = 5.442 × 10^17 × 233 / 6.022 × 10^23
|
||
|
m_Uranium-233 ≈ 21.43 kg
|
||
|
|
||
|
For Plutonium-239:
|
||
|
m_Plutonium-239 = 5.057 × 10^17 × 239 / 6.022 × 10^23
|
||
|
m_Plutonium-239 ≈ 19.97 kg
|
||
|
|
||
|
Therefore, to reach 15,111 megatons of energy,
|
||
|
It would take approximately:
|
||
|
|
||
|
21.04 kg of Uranium-235
|
||
|
21.43 kg of Uranium-233
|
||
|
19.97 kg of Plutonium-239
|