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Fundamentals Of Physics
Found in: Page 1329

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Short Answer

The fission properties of the plutonium isotope Pu239 are very similar to those of U235. The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1.0 kg of pure Pu239 undergo fission?

The released energy is 4.534×1026 MeV.

See the step by step solution

Step by Step Solution

Step 1: Given data

The mass of the plutonium isotope,m = 1 kg

The average energy released per fission,Q = 180 MeV

The molar mass of plutonium, M = 239

Step 2: Determine the formulas to calculate the released energy.

The equation to calculate the number of the atoms in the sample is given as follows.

N=mNAM ...(i)

Here,NA is the Avogadro number (6.022×1023 mol-1)and M is the molar number.

The expression to calculate the released energy is given as follows.

E=NQ ...(ii)

Step 3: Calculate the released energy in the fission.

Calculate the number of the atoms.

Substitute 239 for M , 1 kg for m and 6.022×1023 mol-1 for NA into equation (i).


Calculate the released energy.

Substitute 2.519×1024 for N and 180 MeV for Qinto equation (ii).


Hence the released energy is 4.534×1026 MeV.

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