How do objects emit light?

Question

I recently bumped into a question of how things emit its color. And I've gone through some articles on the internet to be now stuck between two phenomenon: reflection (I'm not sure if its called this way) and spontaneous emission.

According to what I've read, the reason why an apple look red is because it absorbs only certain photons with specific amount of energy (because electrons can only be in some discrete energy level) and reflects (or scatters) the others. So the color "red" that we conceive with our eyes is simply "leftover" photons of the apple's molecules.

That makes sense to me, however, another term came up as I tried to dig further: "spontaneous emission". So when received energy, an electron is excited to a higher energy state, and when that excited electron jumps back to a lower energy level, it emits energy in terms of photons. So the color, in this scenario, that we see from an object is what it absorbed.

So the question here is, if an object reflects lights it cannot absorb (due to reflection) and emits light that it has absorbed (due to spontaneous emission), that means an apple, shined by white light of the sun, would emit all colors within the spectrum of sunlight, however in real life we see apples in red, not in white or rainbow color (unless we paint them). I know that I must be wrong at some points, but I just couldn't figure it out.

Answer 1

An object that emits light would glow in the dark.

If it only reflects light, it returns light that hits it.

Answer 2

Your mistake is assuming that what you see as "red" is a single wavelength of light or a particular combination of wavelengths. In reality, the apple absorbs some colors more efficiently than others, but the spectrum coming off an apple still has all colors of the original spectrum to some degree.

Your brain is adapted to "subtract out" the illuminating spectrum and so perceives an apple as more-or-less the same color regardless of the spectrum it is illuminated with. In other words, humans are very bad spectrographs- you shouldn't assume that two things looking the same means anything at all about the actual spectrums involved.

Answer 3

So the color, in this scenario, that we see from an object is what it absorbed.

No, no and NO. Spontaneous emission does not talk about black body absorption at all. It can be that electron absorbed photon and jumped $E_{0 \to 1}$, then absorbed yet another higher energy photon and jumped further $E_{1\to 2}$, and by spontaneous emission emitting yet higher frequency photon and returning back to ground level by jump $E_{2 \to 0}$. Or it can jump to $E_1$ simply due to energy scattering by phonons (heat dissipation) and then progress to $E_0$ this time emitting photon. Or it may just tunnel to a ground level without emitting anything at all (see quantum tunneling). Or it may stay at the exited state forever,- that's why it's call spontaneous emission, because there's no any guarantee about exact event outcome, just probabilities.

Finally energy absorbed does not necessarily have to be in the form of radiation (photons, light or similar). Our usual light bulbs operate this way. Tungsten rod absorbs electrical power energy, which is then converted to a heat due to Joule heating (metal resistance to a current flow). After that Tungsten emits small part of that energy as visible light waves as a black body (most of absorbed energy will be released as heat/ infrared radiation). What kind of photons bulb will release most,- will depend on Tungsten rod temperature, as is defined by black body radiation law.