Doppler recoil energy

PROBLEM 11.26.1 Compute the Doppler recoil energy ER for 26pe57 and a linear speed vr needed to counteract the Doppler shift. Calculate also the speed of a free recoiling nucleus vr, and comment on why this calculated speed is too big. 

The arguments seen in section 2.3 suggest that resonant y-absorption should decrease at very low temperatures because the Doppler broadening of the y-lines decreases and may even drop below the value of the recoil energy. In his experiments with solid sources and absorbers, however, R.L. Mossbauer ([1] in Chap. 1) observed on the... 

In order to dissipate the recoil energy Mossbauer was the first to use atoms in solid crystal lattices as emitters and also to cool both emitter and absorber. In this way it could be shown that the 7-ray emission from radioactive cobalt metal was absorbed by metallic iron. However, it was also found that if the iron sample were in any other chemical state, the different chemical surroundings of the iron nucleus produce a sufficient effect on the nuclear energy levels for absorption no longer to occur. To enable a search for the precisely required absorption frequency, a scan based on the Doppler effect was developed. It was noted that a velocity of 102 ms-1 produced an enormous Doppler shift and using the same equation (7) it follows that a readily attainable displacement of the source at a velocity of 1 cms-1 produces a shift of 108 Hz. This shift corresponds to about 100 line-widths and provides a reasonable scan width. 

The necessary velocity that would create a Doppler shift corresponding to twice the recoil energy is... 

Figure 10.2. Absorption of y-ray photons by free atoms E = excitation energy E = recoil energy D = line broadening by the Doppler effect.

Here, (...) = J2pn ri. .. ri) is the appropriate combinedquantal and thermodynamic average (over the classical probabilities pn) related with the condensed matter system. M and n(p) are the mass and momentum distribution of the scattering nucleus, respectively, and ior = q2 /2 M is the recoil energy. For convenience, h = 1. Eq. (2) is of central importance in most NCS experiments, since it relates the SCS directly to the momentum distribution. Furthermore, n(p) is related to the nuclear wave function by Fourier transform and therefore, to the spatial localization of the nucleus. It takes into account the fact that, if the scattering nucleus has a momentum distribution in its ground state, the 5-function centered at uor will be Doppler broadened. 

Fundamentals. The energy of a gamma quantum emitted or absorbed by an atomic nucleus usually differs somewhat fi om the actual energy difference between the nuclear states because of recoil and Doppler-broadening induced by thermal movement. At room temperature, the value differs by about 10 to 10 eV from the true energy value vo of the nuclear transition. The recoil energy r and the corresponding momentum p are... 

The second perturbation, namely the thermal motion of emitter nuclei, produce a Doppler-effect broadening of the emission line and causes it to extend in part beyond the energy Eq even though centered at Ey (= Eq—Er). The y-ray energy will be broadened into a distribution by the Doppler-effect energy, E = MvV, which is proportional to the initial velocity, Vx. from the random thermal motion of the atom, and v from the recoil of the nucleus. 

According to eqn (2.20), the change in the kinetic energy of the atom is made up of the Doppler shift of the photon energy and the recoil energy. These two terms play... 

---