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Calculate the energy released, in J, when 1.00 kg of uranium-235 undergoes the following fission process.

\[ \text{n} + ^{235}\text{U} \rightarrow ^{136}\text{I} + ^{96}\text{Y} + 4\text{n} \]

Particle Mass (amu)
- Iodine-136: 135.8401
- Yttrium-96: 95.8629
- Uranium-235: 234.9935
- Neutron: 1.00867

Answer :

The energy released when 1.00 kg of uranium-235 undergoes fission is approximately 2.81 × 10¹² joules.

Initial mass = Mass of Uranium-235 + Mass of Neutron

Final mass = Mass of Iodine-136 + Mass of Yttrium-96 + 4 × Mass of Neutron

Mi = 234.9935 + 1.00867 = 235.00217 amu

Mf = 135.8401 + 95.8629 + 4 × 1.00867

= 135.8401 + 95.8629 + 4.03468

= 235.73768 amu

Mass defect (Δm):

Δm = Mi - Mf

= 235.00217 - 235.73768

= -0.73551 amu

1 amu = 1.660539 × 10⁻²⁷ kg

Mass defect (kg) = -0.73551 × 1.660539 × 10⁻²⁷

= -1.221 × 10⁻²⁸ kg

E = Δm × c²

where c = speed of light = 3 × 10⁸ m/s

E = 1.221 × 10⁻²⁸ kg × (3 × 10⁸ m/s)²

= 1.221 × 10⁻²⁸× 9 × 10¹⁶

= 1.099 × 10⁻¹² J per fission event

Molar mass of U-235 = 235.9935 g/mol

Number of moles in 1 kg = 1000 g / 235.9935 g/mol

= 4.236 mol

Number of atoms = 4.236 mol × 6.022 × 10²³ atoms/mol

= 2.55 × 10²⁴ atoms

Total Energy = 2.55 × 10²⁴ atoms × 1.099 × 10⁻¹² J/atom

= 2.81 × 10¹² J

Note the complete question is:

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