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What must be the relative velocity of a source and an aboserber consisting of freeIr^(191) nuclei to observe the maximum absorption of gamma quanta with energy epsilon=129keV? |
Answer» Solution :For maximum (resonant) absorption, the absorbing nucleus MUST be MOVING with enough SPEED to cancel the momentum of the oncoming photon and have just right enrgy `(epsilon= 129keV)` availiable for transition to excited STATE  Since `deltaE_(gamma)OVERSET~-(epsilon^(2))/(2Mc^(2))` and momentum of the photon is `(epsilon)/(c )`, these condition can be satisfied if the velocity of the nucleus is `(epsilon)/(Mc)=c(epsilon)/(Mc^(2))= 218m//s= 0.218 km//s` |
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