Mössbauer effect

Recoilless emission and recoilless absorption of gamma photons is called Mossbauer effect. It provides us a technique for producing and studying gamma rays whose energies are extremely well defined. Mossbauer effect was discovered by Rudolf Mossbauer in 1958. This effect is applied in the experimental field by the name Mossbauer spectroscopy.

Let us first understand why a nuclear transitions occurs. After a radioactive nucleus undergoes decay (alpha decay, beta decay, k-capture) it many times reaches to the daughter nuclei state but contains too much energy to be in a final stable state. Unlike our electrons orbiting the nucleus, the nucleus itself doesn’t really have such orbits which are easy to visualize. So, let us just imagine it having a particular quantized energy. Now, to be in a stable ground state it must lose this extra energy. It does this by emitting gamma radiation of that energy. This is the process of nuclear transition, where the nucleus transitions from one energy state to a lower one (or higher if it absorbs gamma radiation)

Consider, the stable isotope of iron Fe-57 which has been most often used in Mossbauer effect. Co-57 decays to an excited state of Fe-57 by K-capture. 91% of Fe-57 go to the ground state in two steps. After the emission of a 127keV gamma ray, the nucleus goes in an intermediate state having a half-life of $10^(-7)$s. From this state, the Fe-57 nucleus reaches the ground state by emitting a low energy gamma ray of 14.4 keV. Since the half life of the excited state is known, we can use Heisenberg’s uncertainty principle to estimate the uncertainty in the energy of the gamma emission.

$\Delta E=\frac{\hbar}{\Delta t}=\frac{0.658×10^{-15} eV.s}{10^{-7} s}=6×10^{-9} eV$ 

$\frac{6×10^{-9} eV}{14.4 keV}=4.1667×10^{-13}$

 

Comparing this with the energy of gamma rays, 14.4 keV, we find an uncertainty of about ~ 4 part in $10^(13)$

An extremely small fraction. The energy of an emitted photon is hf when the nucleus transitions from an excited state of energy E to the ground state. So, by conservation of momentum,

$\frac{hf}{c}=MV (p=\frac{h}{\lambda}. \lambda=\frac{c}{f})$

With hf=14.4 keV, if the recoil momentum is given only to the emitting nucleus, we can calculate its recoil energy. For Fe-57, M~$10^(-25)$  kg which gives V~80m/s. This value is low enough for non-relativistic calculations.

$E=\frac{1}{2} MV^2=2×10^{-3} eV$

Now, this same equation applies for recoil energy losses of X-rays , but the photon energy is much less, resulting in lesser energy loss. This is why resonance can be observed with X-rays. This recoil energy is about $3*10^(5)$ times greater than the uncertainty in gamma energy $(\delta E)$ . This reduction in energy seriously affects its chances of getting absorbed by a similar nucleus. There are two ways to overcome this difficulty:

• Doppler Broadening
• To use solid state properties of crystal lattice.

An atom which is embedded in a crystal cannot be treated classically, and since it is bound in a crystal it has only quantized vibrational energy states. These states are widely separated compared to the low recoil energy obtained above implying that the recoil energy would not be able to change the vibrational state of the atom. At a low temperature most of them occupy lowest energy state and effects of the thermal motion become negligible.

So, let us consider one little Co-57 embedded in a crystal lattice. The process occurs and the Fe-57 emits the 14.4 keV energy gamma ray. This is low enough to be insufficient to separate it from the crystal. So now, the recoil energy is not limited to the single atom but is transferred to the group of atoms. This implies that the recoil mass is incredibly large and recoil velocity is almost zero. Thus, the energy of the emitted gamma ray is completely transferred to a similar nucleus kept at the target and resonance absorption occurs.

Mossbauer effect is used as Mossbauer spectroscopy for studying hyperfine interactions, nuclear isomer shift (originates from the Coulomb interaction of the nuclear charge distribution over the radius of the nucleus in its ground state and excited state and electron charge density at the nucleus), nuclear quadrupole splitting (results from the interaction between the EFG at the nucleus and electric quadrupole moment of the nucleus itself), ‘nuclear’ Zeeman effect (nuclear magnetic dipole interacts with an applied magnetic field)

Credit: Nuclear Physics an introduction by S. B. Patel

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