Would a high energy Hydrogen atom start emanating electromagnetic radiation? We know that the total energy of the hydrogen atom is proportional to the inverse of the square of the principal quantum number $n$:
$$E_n \propto -\frac{1}{n^2}$$
So at high quantum numbers the energy spectrum tends towards a continuum.
Shown below, a representation of one of the seven $\text{6f}$ orbitals (courtesy of The Orbitron):

However, due to the Correspondence Principle, high quantum number hydrogen atoms should show wave functions that tend toward Classical orbit-like (instead of orbital) shapes:

This is so, at least according to a video I watched yesterday (unfortunately I don't have the web address)
If this is true, high equantum number orbitals would to become Bohrian in nature and thus emit electromagnetic radiation.
Is this true?
 A: No, you did not break quantum mechanics just because the high energy orbitals are more classical in nature. Think about it like so: suppose an electron is barely bound to the hydrogen atom. Suppose there is an EM field at its ground state with no photons present (we need the field to induce transitions in the hydrogen atom). The electron can now decay to a lower energy state by emitting photons. Because the electron is in such a high quantum level of the hydrogen atom, it can decay by emitting photons of very small energy. To the observer, the emission radiation of a highly excited hydrogen atom gas would approximate a continuum at high energies, and therefore these electrons seem to behave classically at temperatures of about $\sim10-13 ~\text{eV}$, emitting continuous spectrum radiation and supposedly violating quantum mechanics.
However, this classical radiation pattern cannot be continued for arbitrarily low energy. Suppose the electron has been emitting small classical looking packets of energy for a while, becoming more and more strongly bound to the hydrogen atom. At some point, given by the resolution of the experiment and the relevant decay times, the experimentalist is going to start seeing radiation only at certain, clearly discrete wavelengths. The electron has become strongly bound again and has lost its classical character because it came closer to the nucleus, where the transition energies cannot be approximated by a continuum, so they can be deemed classical.
Lesson of the day: Just because something can be approximated as behaving classically in certain regimes of the parameters of the system (high temperature etc.) does not mean it's actually classical. To invoke a theory that doesn't hold in the regime it is to be used to infer results constitutes a logical fallacy.  The correct microscopic theory is decidedly non-classical and it is the one that results like stability of nuclei need to be based on.
A: The answer by @DinosaurEgg addresses the problem of how a neutral hydrogen atom can emit electromagnetic radiation. It does not address the misconceptions in the question.

Would a high energy Hydrogen atom start emanating electromagnetic radiation?


We know that the total energy of the hydrogen atom is proportional to the inverse of the square of the principal quantum number n:

$$E_n \propto -\frac{1}{n^2}$$
This is not the energy of the hydrogen atom, but the difference in energy from the ionization level. Interpreted correctly it means that for high $n$ levels very little energy is needed to ionize the atom.
Nothing to do with high energy.

high energy orbitals would be Bohrian in nature

The italics should be corrected to "high n orbitals"
the high n orbitals are closer to the semiclassical Bohr model
The following  is a non sequitur,

and thus emit electromagnetic radiation.,

It does not follow, as explained in the answer by  @DinosaurEgg. Once an electron is raised to a high $n$ level by some incoming radiation or field, i.e. energy is absorbed by the neutral hydrogen, a soft cascade of photons from the great multiplicity of levels at high n is possible. But it is also possible that it will go directly to a lower level or the ground state, depending on the quantum mechanical probabilities.
Maybe this link will help in studying  the hydrogen atom .
A: A "Bohr orbit" is related to a classical orbit via the correspondence principle, but not all sets of quantum numbers for hydrogen-like atoms correspond to Bohr orbits.  Quantum mechanics is richer than Bohr initially imagined.
Let's consider a hydrogen-like atom with quantum numbers $(n,\ell,m)$. The radial part of the wavefunction is
$$
R_{n\ell}(r) = \sqrt{\text{stuff}}\times e^{-u/2} u^\ell L_{n-\ell-1}^{2\ell+1}(u)
\quad\quad
\text{where } u = \frac{2Zr}{n a}
$$
where $r$ is the radial coordinate, $Z$ the nuclear charge, and $a$ the Bohr radius.
The $L_\alpha^\beta$ are the associated Laguerre polynomials, which are polynomials of order $\alpha$.  So for an orbital with maximal angular momentum $l=n-1$, the Laguerre stuff is just a constant, and the radial wavefunction $R\sim e^{-r} r^\ell$ has a zero at the origin (for nonzero $\ell$) and a single maximum at some finite $r$. For large $n,\ell$ this single peak is narrow and it makes sense to think of the electron as being "radially localized," like a particle in a circular orbit.
Similarly, the spherical harmonics are given by an associated Legendre polynomial in the variable $\cos\theta$, multiplied by an azimuthal phase $e^{im\phi}$.  The extremal Legendre polynomials all have the form
\begin{align}
P_\ell^\ell (x) &= (\text{constant}) (1-x^2)^{\ell/2}
\\
P_\ell^\ell (\cos\theta) &= (\text{constant}) \sin^{\ell}\theta
\end{align}
So for a hydrogen electron where the angular momentum projection is maximized, $|m|=\ell$, the angular probability distribution is strongly peaked at the equator; that peak gets narrower for larger $\ell$.  The complex phase increases linearly as you go around in $\phi$.
Taken together, an electron in a hydrogen-like orbital with maximum angular momentum and maximum angular momentum projection onto the $z$-axis, with quantum numbers $(n,\ell,m) = (\ell+1,\ell,\pm\ell)$, has its probability concentrated in a narrow ring around the equator of the coordinate system, with the phase changing around the ring.
If you include the time-dependent part of the wavefunction, multiplying by $e^{-i\omega t}$, you have a phase change with time which you can use to find a "probability current."  This is the Schrodinger version of a Bohr orbit.
If $\ell$ is large enough that the Bohr orbit is a good description, then the atom can in fact radiate —— not continuously, but by emitting photons and going to orbits with smaller $n$, and eventually to the ground state.
Beware of visualizations like the orbitron, linked in your post. For the case of the $p$-orbitals ($\ell=1$), chemists like to refer to the spatially-oriented $p_x$ and $p_y$, which are rotated versions of $p_z$.  However, $p_z$ has definite $m=0$; the $m=\pm1$ states are linear combinations of $p_x \pm i  p_y$. For the higher $\ell$, there are more choices to make. I think the chemists' approach is to combine wavefunctions with equal-magnitude $m$, so that the combined wavefunctions are real.
A: High energy orbitals can in some respects be represented by classical Bohr orbits but have strictly speaking still to be described my quantum mechanical wave functions, especially when it comes to calculating the transition probabilities to other states. High energy orbitals of all neutral atoms become increasingly hydrogen-like though for increasing n, which makes the transition probabilities easier to calculate.
In any case, radiation is only emitted when the electron makes a transition to a lower level. This is a result of excited atomic states being quantum mechanically unstable (whether n is large or not), not because of the electron radiating as a classical particle. A classically radiating electron would continuously lose energy (producing a continuous spectrum over a wide frequency range in the process) and eventually spiral into the nucleus. This is obviously not observed. One only observes the discrete lines resulting from quantum mechanical transitions from level n to lower states.
The plots below (produced at https://keisan.casio.com/exec/system/1224054805 ) show that, as mentioned above, the wave functions become relatively more sharply peaked with increasing n, but still are wave functions with a continuous spread and do not represent classical orbits.

                   
                   n=3 (l=2)


                   
                   n=10 (l=9)


                   
                   n=100 (l=99)

