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A nucleus of amss 20 u emits a gamma photon of energy 6 MeV . If the emission assume to occur when nuclues is free and at rest then the nulceus will have kintetic energy nearest to (take 1u=1.6xx10^(-27)Kg) |
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Answer» 10 KeV Momentum of photon = momentum of NUCLEUS `(E)/(c )=sqrt(2m(KE))` `(6xx10^(6)xx1.6xx10^(-19))/(3xx10^(8))=sqrt(2xx20xx1.6xx10^(-27)xxKE)` `3.2xx10^(-21)=sqrt(64xx10^(-27)xxKE)` `((3.2)^(2)xx10^(-42))/(64xx10^(-27))=KE` `KE=1.6xx10^(-16)J` `KE=(1.6xx10^(-16))/(1.6xx10^(-19))eV=10^(3)eV=1keV` |
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