1.

A Fe^(57) nucleus emits gamma-rays with an energy of 14.4 keV. Find the relative variation of the energy of gamma-photon due to the recoil of the nucleus. Compare this quantity with the natural width of a spectral line, if the lifetime of a nucleus in the excited state is 1.4xx10^(-7) s.

Answer»


Solution :The energy of the gamma-photon is MUCH HIGHER than that of an ultraviolet photon, the recoil is in this case much greater. Since the energy of a gamma-photon is `epsi_(gamma)=14.4keV`, and the momentum is `p_(gamma)=epsi_(gamma)//c`, the recoil energy is `epsi_(R)=p_(gamma)^(2)//2M=epsi_(gamma)^(2)//2MC^(2)`. Therefore the transition energy is
`epsi=epsi_(gamma)+epsi_(R)=epsi_(gamma)(1+(epsi_(gamma))/(2Mc^(2)))`
The relative change in energy is
`delta=(Deltaepsi)/(epsi_(gamma))=(epsi_(gamma))/(2Mc^(2))`.
The natural relative line width is `delta_(nat)=h//tauepsi_(gamma),"where "tau` is the lifetime of the nucleus in the excited state.


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