This intriguing answer to Can reactions produce microwave or radio wave radiation? says (in part):

An ammonia maser qualifies as an inorganic RF-emitting reaction. As the nitrogen changes position in the $\ce{NH3}$ molecule, ~24 GHz radiation is emitted.

Similarly, organic methanol, $\ce{CH3OH}$ makes a maser found in nature, emitting at ~36.2 GHz and also at ~6.7 GHz

The first link is to The Ammonia Maser in Feynman's Lectures on Physics and it includes a picture of a triangular pyramid with the nitrogen pointing "up" in one case and "down" in the other as an illustration of the transition.

But how does it get there? Does it "tunnel through" the nitrogen triangle, or does the whole molecule flip around? Can we even ask that question? Could we, if for example we labeled the corners with $\ce{{}^{1}H}$ $\ce{{}^{2}H}$ and $\ce{{}^{3}H}$ as a gedankenexperiment to make the two distinguishable?

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    $\begingroup$ The nitrogen atom pierces the triangle of the hydrogen atoms to go on the other side. If you wanted to label something then it should be the hydrogen atoms. There is only one nitrogen atom. But if you arranged $\ce{^1H}$, $\ce{^2H}$, and $\ce{^3H}$ clockwise then you could tell if the nitrogen atom was above or below the triangle of hydrogen atoms. $\endgroup$
    – MaxW
    May 2 '20 at 4:40
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    $\begingroup$ @MaxW yikes $\ce{HN3}$! sorry about that. Because it's such a glaring error I'm going to fix the question accordingly. But how does one know that it pierces the triangle? Hopefully a posted answer will support it's postion. Thanks! $\endgroup$
    – uhoh
    May 2 '20 at 4:43
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    $\begingroup$ Tunnelling occurs from one structure to the other and as a result the levels below the barrier are doubled. The double well potential and some energy levels is shown in this answer, chemistry.stackexchange.com/questions/132449/…. The spectroscopic data shows that in ammonia the barrier is approx 2000 cm$^{-1}$. The N atom is 38 pm out of the plane of the 3 H atoms and the lowest energy transition is 0.97 cm${-1}$. see Swalen, J. D.; Ibers, J. A. J. Chem. Phys. 1966, 36, 1914 - 1918. $\endgroup$
    – porphyrin
    May 2 '20 at 7:57
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    $\begingroup$ if I remember correctly from my undergraduate lectures, all saturated N centres (ammonia & amines) flip continually back and forth at a very high rate. This is the reason that a tertiary amine with 3 differing R groups is not chiral. This was always explained by the sp3 orbital (that the lone pair on the Nitrogen atom occupies) has a tiny probability of being on the opposite side of the N atom. This allows the N centre to invert with a low energetic barrier. $\endgroup$
    – MNB
    May 2 '20 at 21:55
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    $\begingroup$ @uhoh You should consider the reduced mass of the system as the mass that tunnels. In a simplified picture, the hydrogen atoms are considered to move simultaneously with respect to the nitrogen atom and the reduced mass is given by $\mu_\text{red}=3m_\text{N}m_\text{H}/(m_\text{N}+3m_\text{H})\approx 2.5$ u, which is a relatively light and can easily tunnel. Another proof that the ammonia really tunnels is by comparing the observed splitting of NH$_3$ with that of ND$_3$ and NT$_3$ which follows exactly what you expect for tunneling. $\endgroup$
    – Paul
    May 4 '20 at 18:46

Let me expand a little on my earlier comment. If we only consider the inversion of the ammonia molecule (we ignore rotation and vibrations other than the inversion mode), the Schrödinger equation is given by $$ -\frac{\hbar^2}{2\mu_\text{red}}\frac{\text{d}^2\psi{z}}{\text{d}z^2}+V(z)\psi(z)=E\psi(z), $$ with $\mu_\text{red}$ the reduced mass and $z=d_\text{NH}\sin{\alpha}$, where $d_\text{NH}$ is the NH bond distance and $\alpha$ the so-called umbrella angle. The potential V(z) is shown in the link given by porphyrin in one of his comments. Although the reduced mass does, in principle, depend on $\alpha$ (see Eq. 4 of this paper), the dependance is weak and we may consider that the hydrogen atoms move simultaneously with respect to the nitrogen atom. In this case $$ \mu_\text{red}=\frac{3m_\text{H}m_\text{N}}{3m_\text{H}+m_\text{N}}\approx2.5\text{ u}. $$ The tunneling does not change the center of mass (c.o.m.) of the molecule, so while the nitrogen moves up, the plane of hydrogen atoms moves down to ensure the stationary c.o.m. (and vice versa). I made the following animated gif at some point during my PhD (sorry for the watermark, but I've had some negative experiences in the past):

enter image description here

The first paper - to my knowledge - that treats the tunneling of ammonia appeared already in 1932. The first application of QM tunneling was done by Gamow appeared 4 years prior to the ammonia work and solved the mystery of the huge range of observed lifetimes of alpha emitters (20 orders of magnitude).

In the 1932 paper, Dennison and Uhlenbeck also derive the magnitude of the tunneling splitting as a function of the reduced mass $$ E_\text{inv}=\frac{h\nu_0}{\pi}\exp\left \{-\frac{2\pi}{h}\int_\limits{-z_0}^{z_0}\left [2\mu_\text{red}\left (V(z)-E \right ) \right ]^{1/2} \right \}, $$ where $\pm z_0$ are the equilibrium positions corresponding to the potential minima, $\nu_0$ is the vibration frequency in one of the two wells and $E$ is the energy of the system with respect to the potential minimum. This expression is derived from the WKB approximation and it basically is the product of the tunneling probability with the frequency with which the reduced mass 'hits' the barrier. As can be seen from the formula, the tunneling frequency scales exponentially with the square root of the reduced mass, which is also observed experimentally. The following picture is taken from my thesis and basically a copy of Fig. 2 of this paper.

enter image description here

If the molecule would simple rotate instead of tunnel, the change in the energy splitting would be inversely proportional to the reduced mass (via the moment of inertia) and this is clearly not observed.

  • $\begingroup$ This is of course an excellent answer! I'm off to the library today to get these... $\endgroup$
    – uhoh
    May 5 '20 at 22:42
  • $\begingroup$ hmm... I'm not sure what happened but I don't seem to have followed through on this, So I'm off to the library today again @@ to get these... ;-) $\endgroup$
    – uhoh
    Aug 3 '20 at 1:46

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