Can photons of visible light lose energy and become sub-visible?

[royp]

TL;DR Summary

When a red-hot metal allowed to cool down in a room what happens to the 'red' visible photons: Do they become low energy, low frequency, sub-visible-spectra photons?

Let me clarify this by a thought experiment:

Imagine, a heated, red-hot (emitting only monochromatic red light) metal bar is brought inside a dark room. The room is practically
insulated and the metal bar is glowing in the dark - emitting 'red photons' of visible light. Eventually, the bar will
cool down, the 'steady state' average temparature is achieved inside the room and the overall room temparature is slightly up.

As there are no more red glow (photons), have those photons of visible light 'degenerated' into lower energy(frequency) of
sub-visible energy spectrum? This is my fundamental question.

But in this connection, I also would like to add the following: Niels Bohr in his phenomenal explanation of Hydrogen
(absorption) spectra, postulated that electrons are going to 'outer' (higher energy) orbit by absorbing the photons
(of certain colour) and eventually coming back to its original state/orbit by releasing the photon; the same energy
photon as absorbed earlier - That is my understanding.

Or is it that the electrons often absorb the high energy photons and release low energy photons later?

Otherwise, how do we explain the metal bar losing its 'glow'?

Many thanks in advance for your responses!

 

[PeterDonis]

As there are no more red glow (photons), have those photons of visible light 'degenerated' into lower energy(frequency) of
sub-visible energy spectrum?

No, they've been absorbed by the walls of the room. The infrared photons present in the room once the steady state is achieved are new photons, emitted by the bar (and the walls of the room).

electrons are going to 'outer' (higher energy) orbit by absorbing the photons
(of certain colour) and eventually coming back to its original state/orbit by releasing the photon; the same energy photon as absorbed earlier - That is my understanding.

Yes. However, this isn't why the metal bar emits red visible light photons when it's red hot, for two reasons.

First, the bar didn't originally get heated up by absorbing visible light photons. You don't heat up a metal bar to red hot by shining red light at it. You heat it up by putting it in a fire. That's a different process.

Second, when the bar is red hot, it's not because electrons in individual atoms in the bar are in higher energy levels. It's because the atoms in the bar as a whole are jiggling around in place with more kinetic energy, and to some extent because free electrons in the metallic energy band in the bar have more kinetic energy. The visible red photons in the bar are emitted as the atoms and free electrons lose kinetic energy; they aren't emitted from individual electron transitions to lower energy levels in atoms.

how do we explain the metal bar losing its 'glow'?

Because its temperature goes down as it emits light, and as its temperature goes down, the atoms in it have less kinetic energy, so they emit photons of lower energy.