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PhysicsModern PhysicsNuclear physicsEasy2 minPYQ_2025

PhysicsEasysingle choice

Energy released when two deuterons $(_{1} H^{2})$ fuse to form a helium nucleus $(_{2} He^{4})$ is :
(Given : Binding energy per nucleon of $_{1} H^{2} = 1.1 MeV$ and binding energy per nucleon of $_{2} He^{4} = 7.0 MeV$ )

Options:

Answer:

Solution:

$E_{B} = BE_{reactant} - BE_{product} = 1.1 \times 2 + 1.1 \times 2 - 7 \times 4 = - 23.6 MeV = Q = 23.6 MeV$

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Nuclear physics

⏱ 2mℹ️ Source: PYQ_2025

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