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JEE Advanced2019PhysicsNuclear PhysicsActual

Suppose a R a 88 226 nucleus at rest and in ground state undergoes α -decay to a R n 86 222 nucleus in its excited state. The kinetic energy of the emitted α particle is found to be 4.44 M e V . R n 86 222 nucleus then goes to its ground state by γ -decay. The energy of the emitted γ -photon is _______ k e V , [Given: atomic mass of 88 226 R a = 226.005   u , atomic mass of 86 222 R n = 222.

Correct answer

0

Step-by-step solution

⇒ Mass defect ∆ m = 226.005 - 222.000 - 4.000 R a 88 226 → α - d e c a y R n + 2 4 H e + γ 86 222 = 0.005 a m u ∴ Q v a l u e = 0.005 × 931.5 = 4.655 M e V Also K . E α K . E R n = m R n m α ⇒ K . E R n = m α m R n . K . E α = 4 222 × 4.44 = 0.08 M e V ∴ Energy of γ - P h o t o n = 4.655 - 4.44 + 0.08 E γ = Q - K E α + K E R n = 0.135 M e V = 135 K e V

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