In the process called fluorescence, some substances absorb photons of one kind of radiation (e.g) ultraviolet radiation), and give out photons of a different kind (e.g visible light). When infrared radiation falls on these materials, visible light is not emitted. Why not?
If the energy to be given out must be less than that of the incident photons, then the wavelength of the light being readmitted will not be in the visible range. Think about that.
In the process of fluorescence, certain substances absorb photons of one type of radiation, such as ultraviolet (UV) radiation, and then emit photons of a different type, such as visible light. This phenomenon occurs because the energy from the absorbed photons excites the electrons within the substance, causing them to move to higher energy levels. When these excited electrons return to their original energy levels, they release the excess energy in the form of photons.
Now, when infrared (IR) radiation falls on these fluorescent materials, visible light is not emitted. This is because the energy of the IR photons is typically lower than the energy required to excite the electrons to higher energy levels. In order for visible light to be emitted, the energy of the emitted photons must be lower than the energy of the absorbed photons. Since the energy of IR radiation is generally lower than that of visible light, it cannot provide sufficient energy to excite the electrons to the necessary energy levels that would lead to the emission of visible light.
To summarize, the reason why visible light is not emitted when IR radiation falls on fluorescent materials is that the energy of IR photons is insufficient to excite the electrons in these materials to the required energy levels for visible light emission.