The following equation represents a fusion reaction: 1 H 2 + 1 H 3 → 2 H e 4 + 0 n 1 + Q ; where Q is the energy released. Given: mass of 1 H 2 atom = 2 . 014102 u , mass of H 3 1 atom = 3 . 0 16050 u , mass of H 2 e 4 atom = 4 . 002603 u and mass of neutron n 1 0 = 1 . 008665 u . Calculate the value of Q .

In the given reaction: 1 H 2 + 1 H 3 → 2 H e 4 + 0 n 1 + Q Mass of reactants = m ( 1 H 2 ) + m ( 1 H 3 ) m i = 2 . 0141202 + 3 . 016049 = 5 . 030151 u Mass of products = m ( 2 H e 4 ) + m ( 0 n 1 ) = 4 . 002603 + 1 . 008665 = 5 . 011268 u Mass defect Δ m = 5 . 030151 u − 5 . 011268 u = 0 . 018883 u Energy associated with 1 a . m . u = 931 MeV Thus, Energy released = 0 . 018883 × 931 MeV = 17 . 58 M e V

The following equation represents a fusion reaction: 1 H 2 + 1 H 3 → 2 H e 4 + 0 n 1 + Q ; where Q is the energy released. Given: mass of 1 H 2 atom = 2 . 014102 u , mass of H 3 1 atom = 3 . 0 16050 u , mass of H 2 e 4 atom = 4 . 002603 u and mass of neutron n 1 0 = 1 . 008665 u . Calculate the value of Q .

In the given reaction: 1 H 2 + 1 H 3 → 2 H e 4 + 0 n 1 + Q Mass of reactants = m ( 1 H 2 ) + m ( 1 H 3 ) m i = 2 . 0141202 + 3 . 016049 = 5 . 030151 u Mass of products = m ( 2 H e 4 ) + m ( 0 n 1 ) = 4 . 002603 + 1 . 008665 = 5 . 011268 u Mass defect Δ m = 5 . 030151 u − 5 . 011268 u = 0 . 018883 u Energy associated with 1 a . m . u = 931 MeV Thus, Energy released = 0 . 018883 × 931 MeV = 17 . 58 M e V

E M B I B E

ISC Physics Class 12, Part-2>Chapter 17 - Nuclear Fission and Nuclear Fusion : Sources of Energy>QUESTIONS>Q 10

HARD

12th ICSE

IMPORTANT

Earn 100

The following equation represents a fusion reaction: $_{1} H^{2} +_{1} H^{3} \to_{2} H e^{4} +_{0} n^{1} + Q$ ; where $Q$ is the energy released. Given: mass of $_{1} H^{2}$ atom $= 2.014102 u$ , mass of ${}_{1}H^{3}$ atom $= 3.0 16050 u$ , mass of ${}_{2}H e^{4}$ atom $= 4.002603 u$ and mass of neutron ${}_{0}n^{1} = 1.008665 u$ . Calculate the value of $Q$ .

Important Questions on Nuclear Fission and Nuclear Fusion : Sources of Energy

HARD

12th ICSE

IMPORTANT

ISC Physics Class 12, Part-2>Chapter 17 - Nuclear Fission and Nuclear Fusion : Sources of Energy>For DIFFERENT COMPETITIVE EXAMINATIONS>Q 1

Consider the fission reaction:

$_{92} U^{236} ⟶ X^{117} + Y^{117} + 2_{0} n^{1}$ . The binding energy per nucleon of $U, X$ and $Y$ is $7.6 MeV, 8.5 MeV$ and $8.5 MeV$ respectively. The total energy liberated is about: