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PhysicsModern PhysicsNuclear physicsEasy2 minPYQ_2022

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Nucleus $A$ is having mass number $220$ and its binding energy per nucleon is $5.6 MeV$ . It splits in two fragments $B$ and $C$ of mass numbers $105$ and $115$ . The binding energy of nucleons in $B$ and $C$ is $6.4 MeV$ per nucleon. The energy $Q$ released per fission will be:

Options:

Answer:

Solution:

The total binding energy of nucleus $A$ is,

$E_{A} = 220 \times 5.6 = 1232 MeV$

The total binding energy of $B$ and $C$ combined is,

$E_{B} + E_{C} = 105 \times 6.4 + 115 \times 6.4 = 1408 MeV$

Therefore, the energy released per fission is,

$Q = E_{B} + E_{C} - E_{A} = 1408 - 1232 = 176 MeV$

Stream:JEESubject:PhysicsTopic:Modern PhysicsSubtopic:Nuclear physics

⏱ 2mℹ️ Source: PYQ_2022

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