Environmental Science: Numerical Questions – Energy (Nuclear Fission)

Dr. Manishika Jain- Join online Paper 1 intensive course. Includes tests and expected questions.

Nuclear Fission

Q 1) A typical fission reaction involving an atom of 92U235 leads to a mass defect =0. 37×10−27 kg. How much energy is going to be released by 1.0 g of 92U235?

(1) 3.33 × 10−11 MJ

(2) 2.23 × 1012 MJ

(3) 2.56 × 104 MJ

(4) 8.53 × 104 MJ

Answer: Option (4) 8.53 × 104 MJ

Mass Defect = 0. 37×10−27 kg

Amount of 92U235 = 1 gm

E = mc2

E = 0.37∗ 𝟏𝟎−𝟐𝟕 ∗ (𝟑∗𝟏𝟎𝟖 )𝟐

E = 0.37 ∗ 𝟏𝟎−𝟐𝟕 ∗ 𝟗∗ 𝟏𝟎𝟏𝟔 = 3.33 * 10-11

Number of Atoms in 5 gm of 92U235 = nA

= 1/235 ∗ 6.023 ∗ 1023

= 0.0042 ∗ 6.023 ∗ 1023

= 0.026 ∗ 1023

Total Energy = Energy of 1 atom * Number of Atoms

= 3.33 ∗ 10−11 ∗ 0.026 ∗ 1023

= 0.0865 * 1012 = 8.65 * 104 MJ

Q 2) In a nuclear fission reaction involving 235U92 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 235U92? (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 92U235 = 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 = mc2

Total Energy Released from 5 gm of 92U = nA * mc2

n = number of moles

A = Avogadro’s number (6.023 * 1023)

= 437.19 GJ