Question

The fission properties of 23994Pu are very similar to those of 23592U. The average energy released per fission is 180 MeV. How much energy, in MeV, is released if all the atoms in 1 kg of pure 23994Pu undergo fission?

Answer 1

4.536×10²⁶ MeV released if all the atoms in 1 kg of pure  $^{239} Pu _{94}$  undergo fission

Explanation:

Number of atoms in 1  g of pure $^{239} Pu _{94}$   =  (The Avogadro constant  / 239 )  

The Avogadro constant = 6.023×10²³

Number of atoms in 1  g of pure $^{239} Pu _{94}$  =  (6.023×10²³ / 239 )  

1 kg = 1000 gm

Number of atoms in 1 kg of pure $^{239} Pu _{94}$ = (6.023×10²³ / 239 )   ×1000

Number of atoms in 1 kg of pure $^{239} Pu _{94}$  =  =2.52×10²⁴

Average energy released in fission is 180MeV

=>  the total energy released = Average energy * Number of atoms

= 2.52×10²⁴ × 180 MeV

= 4.536×10²⁶ MeV

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Answer 2

Number of atoms in 1 kg of pure Pu=

$\bf \frac{6.023 \times {10}^{23} }{239} \times 1000 = 2.52 \times {10}^{24}$

As average energy released in fission is 180MeV, the total energy released is

⇒2.52×10

24 ×180MeV

⇒4.53×10 ^26 MeV.