1.

What should be the relative speed of approach of a source and an absorber consisting of free iron atoms for the resonance absorption of gamma-rays to take place? The energy of the photon is specified in the previous problem.

Answer»


Solution :The absorption of a gamma-PHOTON by a free nucleus results in it acquiring a momentum equal to that of the photon and, consequently, a kinetic energy `epsi_(R)=p_(gamma)^(2)//2M=epsi_(gamma)^(2)//2Mc^(2)`. Therefore the energy of the ABSORBED photon is `epsi_(gamma)^(ab)=EPSI+epsi_(R),"where "epsi=epsi_(gamma)+epsi_(R)` is the energy of the transition, and `epsi_(gamma)` is the energy of the emitted PHOTONS. Hence
`epsi_(gamma)^(ab)=epsi_(gamma)+2epsi_(R)=epsi_(gamma)(1+epsi_(gamma)/(Mc^(2)))`
This distorts the resonance absorption. If we start bringing the source and the absorbing substance closer together, the gamma-photon.s energy increases because of the Doppler effect: `epsi_(gamma).=epsi_(gamma)sqrt((1+beta)/(1-beta))~~epsi_(gamma)(1+beta)`. At a certain speed we obtain `epsi_(gamma).=epsi_(gamma)^(ab)`, and resonance absorption sets in. We have
`epsi_(gamma)(1+epsi_(gamma)/(Mc^(2)))=epsi_(gamma)(1+beta),"giving "beta=(epsi_(gamma))/(Mc^(2)),andv=betac=epsi_(gamma)/(Mc)`


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