Recoil-Free Events

The Mössbauer effect occurs when atoms are in a solid lattice or matrix. The chemical binding energies in solids (1-10 $\unskip\,\mathrm{eV}$) are much greater than free atom recoil energies, $E_{R}$. The mass, $M$, recoiling then becomes effectively that of the entire crystal, which can be of the order of $10^{15}$ greater than a single atom. It can be seen from equations 2.1 and 2.2 that $E_{R}$ and $E_{D}$ will now be negligible in this case.

However, although the nucleus is bound within the lattice it is still free to vibrate. The recoil energy can still be transferred to the lattice as a quantised lattice vibration, or phonon. If the recoil energy is less than the lowest quantised vibrational mode then a recoil-free event will occur. The probability of such an event is governed by the recoil-free factor, $f$, which is given as

$$f=exp\left( \frac{ -E_{\gamma}^{2}\langle{}x^{2}\rangle{} }{ \hbar{}c^{2} } \right)$$ (2.4)

where $\langle{}x^{2}\rangle{}$ is the mean square vibrational amplitude of the emitting or absorbing nucleus.[5] It can be seen that $f$ decreases exponentially with the square of the gamma ray energy; this is why the Mössbauer effect is only detected in isotopes with a very low lying excited state. The other dependent factor, $\langle{}x^{2}\rangle{}$, is a function of both the binding strength and temperature. The optimum $f$ factor, and hence the best signal/noise ratio, is obtained for isotopes with very low lying excited states at temperatures well below their Debye Temperature, $\theta{}_{D}$. A good example is $^{57}$Fe, with a Mössbauer gamma ray energy of 14.41 $\unskip\,\mathrm{keV}$ and a $\theta{}_{D}$ of 470 $\unskip\,\mathrm{K}$, allowing strong signals to be recorded at room temperature.