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PhysicsModern PhysicsAtomic structureMedium2 minQB

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A stationary nucleus of mass 25 u emits a photon of energy 3.05 MeV. The recoil energy of the nucleus is (Take $1 u = 931 MeV$ )

Answer:

Solution:

$E_{recoil} = \frac{E _{γ}^{2}}{2 M c ^{2}} = \frac{( 3.05 ) ^{2}}{2 \times 25 \times 931} MeV = 2.0 \times 1 0^{- 4} MeV = 0.2 keV$ .

Subject:PhysicsTopic:Modern PhysicsSubtopic:Atomic structure

⏱ 2mℹ️ Source: QB

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