# Environmental Science: Numerical Questions – Energy (Nuclear Fission)

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## Nuclear Fission

Q 1) A typical fission reaction involving an atom of _{92}U^{235} leads to a mass defect =0. 37×10−27 kg. How much energy is going to be released by 1.0 g of _{92}U^{235}?

(1) 3.33 × 10^{−11} MJ

(2) 2.23 × 10^{12} MJ

(3) 2.56 × 10^{4} MJ

(4) 8.53 × 10^{4} MJ

Answer: Option (4) 8.53 × 10^{4} MJ

Mass Defect = 0. 37×10−27 kg

Amount of _{92}U^{235} = 1 gm

E = mc^{2}

E = 0.37∗ 𝟏𝟎^{−𝟐𝟕} ∗ (𝟑∗𝟏𝟎^{𝟖} )^{𝟐}

E = 0.37 ∗ 𝟏𝟎^{−𝟐𝟕} ∗ 𝟗∗ 𝟏𝟎^{𝟏𝟔} = 3.33 * 10^{-11}

Number of Atoms in 5 gm of _{92}U^{235} = nA

= 1/235 ∗ 6.023 ∗ 10^{23}

= 0.0042 ∗ 6.023 ∗ 10^{23}

= 0.026 ∗ 10^{23}

Total Energy = Energy of 1 atom * Number of Atoms

= 3.33 ∗ 10^{−11} ∗ 0.026 ∗ 10^{23}

= 0.0865 * 10^{12} = 8.65 * 10^{4} MJ

Q 2) In a nuclear fission reaction involving ^{235}U_{92} and a slow neutron, the mass defect is found to be 0.223 U. How much energy will be released from 5.0 gram of ^{235}U_{92}? (1U = 1.66 × 10^{−27} kg)

(1) 426.7 GJ

(2) 85.3 GJ

(3) 170.6 GJ

(4) 42.6 GJ

Answer: Option (1) 426.7 GJ

Mass Defect = 0.223 u

Amount of _{92}U^{235} = 5 gm

1u = 1.66 x 10^{-27} kg

Mass Defect = 0.223 * 1.66 x 10^{-27} (As, 1u is equal to 1.66 x 10^{-27} kg)

= 0.3791 x 10^{-27} kg

E = mc^{2}

Total Energy Released from 5 gm of ^{92}U = nA * mc^{2}

n = number of moles

A = Avogadro’s number (6.023 * 10^{23})

= 437.19 GJ

-Mayank