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PhysicsModern PhysicsMiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

Consider the nuclear reaction $X^{200} \to A^{110} + B^{80}$ . If the binding energy per nucleon for $X$ , $A$ and $B$ are $7.4 MeV$ , $8.2 MeV$ and $8.1 MeV$ respectively, then the energy released in the reaction is

Options:

Answer:

Solution:

For $X$ , energy $= 200 \times 7.4 = 1480 MeV$

For A, energy $= 110 \times 8.2 = 902 MeV$

For B, energy $= 80 \times 8.1 = 648 MeV$

Therefore energy released

$= (902 + 648) - 1480$

$= 1550 - 1480 = 70 MeV$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Modern PhysicsSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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