A nucleus of mass M emits γ-ray photon of frequency v. The loss of internal energy by the nucleus is: [Take c as the speed of electromagnetic wave]

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Correct answer: (D) hv1+hv/(2Mc^2). Energy of γ ray E_(γ)=hv Momentum of γ ray P_(γ)=h/(λ)=hv/c Total momentum is conserved. P→_(γ)+P→_Nu=0 where P→_Nu= momentum of decayed nuclei ⇒P_(γ)=P_Nu ⇒hv/c=P_Nu ⇒KE of nuclei =1/2Mv^2=PNu^22M=1/2Mhvc^2 Loss in internal energy =E_(γ)+KE_Nu =hv+1/2Mhvc^2 =hv1+hv/(2Mc^2)

A nucleus of mass $M$ emits $γ$ -ray photon of frequency $v$ . The loss of internal energy by the nucleus is: [Take $c$ as the speed of electromagnetic wave]

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