A nucleus of mass 20u emits a photon of energy 6 MeV. If the emission assume to occur when nucleus is free and at rest, then the nucleus will have kinetic energy nearest to (take 1u=1.6 10^-27 kg )

Question

Accepted Answer

Correct answer: B. (b) Mass of nucleus, M=20u E_ =6 MeV=6 1.6 10^-13 J Using momentum conservation principle, 0= Mv+ E_ c ∴ v=- E_ Mc=- 6 1.6 10^-1320 1.6 10^-27 3 10^8 =-10^5 m s^-1 K.E. of nucleus, K= 12Mv^2 = 12 20 1.6 10^-27 (10^5 )^2=16 10^-17 J = 16 10^-171.6 10^-16 keV=1 keV

$A nucleus of mass 20 u emits a γ photon of energy 6 MeV . If the emission assume to occur when nucleus is free and at rest, then the nucleus will have kinetic energy nearest to (take 1 u = 1.6 \times 10^{- 27} kg)$

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