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NEETPhysicsCenter of Mass, Momentum and Collision

A stationary nucleus of mass number A undergoes alpha decay by emitting an alpha particle of mass 4 . If the total kinetic energy released in the reaction is Q , what is the kinetic energy of the recoiling daughter nucleus?

Options

  1. A4Q A
  2. B(A-4)Q A
  3. C4Q A+4
  4. DQ 2

Correct answer

A. 4Q A

Step-by-step solution

Let the momentum of the emitted alpha particle be p . By the law of conservation of linear momentum, the recoiling daughter nucleus must have an equal and opposite momentum of magnitude p . The mass of the alpha particle is 4 and the mass of the daughter nucleus is (A-4) . The kinetic energy of the alpha particle is K_ = p^2 2(4) and the kinetic energy of the daughter nucleus is K_D = p^2 2(A-4) . The total kinetic energy released is Q = K_ + K_D = p^2 2(4) + p^2 2(A-4) . Simplifying the expression for Q : Q = p^2

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