The fission properties of Pu239 are very similar to those of 92U235. How…

The fission properties of Pu239 are very similar to those of 92U235. How much energy (in MeV), is released if all the atoms in 1 g of pure 94Pu239 undergo fission? The average energy released per fission is 180 MeV.

Ans.

To find the energy released when all the atoms in 1 g of pure 94��23994Pu239 undergo fission, we need to calculate the number of atoms in 1 g of 94��23994Pu239 and then multiply it by the average energy released per fission.

First, we need to find the number of moles of 94��23994Pu239 in 1 g using its molar mass.

The molar mass of 94��23994Pu239 is approximately $239\text{g/mol}$.

So, the number of moles of 94��23994Pu239 in 1 g is:

Number of moles=MassMolar mass=1 g239 g/molNumber of moles=Molar massMass​=239g/mol1g​

Now, we use Avogadro’s number (6.022×1023 atoms/mol6.022×1023atoms/mol) to find the number of atoms:

Number of atoms=Number of moles×6.022×1023 atoms/molNumber of atoms=Number of moles×6.022×1023atoms/mol

Once we have the number of atoms, we can find the total energy released:

$$\text{Total energy released}=\text{Number of atoms}\times \text{Energy per fission}$$

Given that the average energy released per fission is 180 MeV, we can use this value to find the total energy released.

Let’s perform the calculations:

Number of moles=1 g239 g/mol≈4.18×10−3 molNumber of moles=239g/mol1g​≈4.18×10−3mol

Number of atoms=4.18×10−3×6.022×1023≈2.515×1021 atomsNumber of atoms=4.18×10−3×6.022×1023≈2.515×1021atoms

Total energy released=2.515×1021×180 MeVTotal energy released=2.515×1021×180MeV

Total energy released≈4.5287×1023 MeVTotal energy released≈4.5287×1023MeV

Therefore, if all the atoms in 1 g of pure 94��23994Pu239 undergo fission, approximately 4.5287×10234.5287×1023 MeV of energy would be released.