Mossbauer effect explanation I need to understand Mossbauer effect. Is there any simple explanation? So far I know, when an atomic nucleus emits a gamma-ray photon, the nucleus must recoil to conserve linear momentum.
Consequently there is a change of frequency of the radiation due to the movement of the source the Doppler effect.
What it is good for, this recoils absorption, or emission?
What are the  consequences?
What is so special about Mossbauer spectroscopy?
 A: Starting from

So far I know, when an atomic nucleus emits a gamma-ray photon, the nucleus must recoil to conserve linear momentum.

is to begin with a classical expectation.
Instead say, "When a system emits a gamma ray the system must recoil", but allow that the system could be the single atomic nucleus or if the atom is part of a solid it could be the whole lump of solid matter.
That wouldn't be possible in classical physics because the recoil would have to be transmitted sequentially from one atom to another by inter-atomic forces, but in quantum mechanics the final state in which the whole crystal recoils coherently is consistent with the initial quantum numbers of the system so. But the Totalitarian Principle, holds that outcomes not forbidden must be possible, so outcomes with coherent recoil must happen. It remains to compute the rate (which is difficult), and determine how that effects the system (recoil energy must be deducted from the photon energy).
Getting the photon energy
Compute the final state for a single photon emitted from a system with mass $M^*$ resulting in mass $M < M^*$ in the rest frame of the initial state, by conserving energy and momentum
\begin{align}
M^* c^2 
&= E_M + E_\gamma \\
&= \left(Mc^2 + \frac{{P_M}^2}{2M}\right) + E_\gamma \tag{1}\\
0 &= P_M - p_\gamma\\
p_\gamma &= P_M \tag{2} \;.
\end{align}
Where I have used the Newtonian kinetic energy for the slowly recoiling system.
Now, from (2) we find $P_M = p_\gamma = \frac{E_\gamma}{c}$, so we can re-write (1) as
\begin{align}
\frac{1}{2Mc^2} E_\gamma^2 + E_\gamma - (M^* - M)c^2 &= 0 \\
E_\gamma^2 + 2Mc^2E_\gamma - 2M(M^* - M)c^4 &= 0 \\
\end{align}
which leads to
\begin{align}
E_\gamma 
&= \frac{1}{2} \left[ -2Mc^2 \pm \sqrt{4M^2 c^4 + 8 M (M^* - M)c^4} \right] \\
&= Mc^2 \left[ -1 \pm \sqrt{1 + 2 \frac{\Delta M}{M}} \right]\\
\end{align}
Expanding the square root about zeroto three terms gives
\begin{align}
E_\gamma 
&\approx Mc^2 \left[ \frac{\Delta M}{M} - \frac{1}{2}\left(\frac{\Delta M}{M}\right)^2 \right] \\
&= \Delta M c^2 \left[1 - \frac{\Delta M}{2M} \right] \;.
\end{align}
The result is that the larger the system mass, the closer the photon energy is to the total available energy. (Keep in mind that we implicitly assumed $\Delta M \ll M$ when we chose the Newtonian kinetic energy at the start.)
Most sources then go on to make the further approximation of replacing $\Delta M$ by the gamma energy itself.
