A free sationary `Ir^(191)` nucleus with excitation energy `E=129 keV` passes to the ground state, emitting a gamma quantum. Calculate the fractional change of gamma quanta energy due to recoil of the nucleus.

A free sationary `Ir^(191)` nucleus with excitation energy `E=129 keV`

1.	A free sationary `Ir^(191)` nucleus with excitation energy `E=129 keV` passes to the ground state, emitting a gamma quantum. Calculate the fractional change of gamma quanta energy due to recoil of the nucleus.
Answer» With recoil neglected, the `gamma` quantum will have `129keV` enrgy. To a first approximation, its momenturm will be `129keV//c` and the energy of recoil will be `((0.129)^(2))/(2xx191xx931)MeV= 4.18xx10^(-8)MeV` In the next approximation we therefore write `E_(gamma)overset~-.129-8.2xx10^(-8)MeV` Therefore `(deltaE)/(E_(gamma))=3.63xx10^(-7)`

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A free sationary `Ir^(191)` nucleus with excitation energy `E=129 keV` passes to the ground state, emitting a gamma quantum. Calculate the fractional change of gamma quanta energy due to recoil of the nucleus.

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