It is definitely possible, although many nanopowders do not emit in solid state (i.e. without being dispersed in a medium) because of aggregation induced quenching. A good example of this is carbon nanodots, very emissive in solution, yet quenched in solid-state. Most dye molecules have the same behaviour. If your sample emits in powder form, you can record its emission in "front-face" geometry within a spectrofluorometer. In this geometry, the illuminated portion of the sample (hosted within a suitable sample holder, such as an aluminum cilinder) is facing (45 degree angle) the detector. Some emission will be collected from the detector, i.e. you can get the spectrum you're looking for.

Regards

Fabrizio

It is definitely possible, although many nanopowders do not emit in solid state (i.e. without being dispersed in a medium) because of aggregation induced quenching. A good example of this is carbon nanodots, very emissive in solution, yet quenched in solid-state. Most dye molecules have the same behaviour. If your sample emits in powder form, you can record its emission in "front-face" geometry within a spectrofluorometer. In this geometry, the illuminated portion of the sample (hosted within a suitable sample holder, such as an aluminum cilinder) is facing (45 degree angle) the detector. Some emission will be collected from the detector, i.e. you can get the spectrum you're looking for.

Regards

Fabrizio

Dark brown appearance infers that your material is somewhat less absorbing in the orange-red range. If your emission line or lines fall into this range and sensitivity of your spectrometer is sufficiently high, you can get a measurable signal. The efficiency of emission, however, won't be high so I would doubt that your material would be useful from a practical point of view, if you intend it to be a phosphor in a powder state.

Many years ago, I had synthesized dark brown GaN powders with Zn-doping. The powders emitted blue when excited by 325 nm He-Cd laser beam. Anyway, the color of the powder depends on the syntesis temperatrue which influences the desoprtion of the carbon from the precusors. At low synthesis temperature, the products look black; at high synthesis temperatrue, the products look grey, while it looks dark brown at medium temperature.

As long as the powder isn't black it can emit, even in the visible spectral range.

However, quantum efficiency won't be good as the material will absorb its own emission partially.

High quality luminescent materials should have close to zero absorption in the range where they emit. But still, even if absorption is high, they can show emission, albeit weak.

As Fabrizio replied, some materials form aggregates in solid state owing to their molecular nature. I used to study some BODIPY dyes that were dark brown or even very dark green in solids by reason of association caused by F-H intermolecular bond formation. The powders emitted in the 650-900 nm range. However, in diluted solutions the dyes demonstrated efficient green fluorescence.

It is also possible to observe luminescence from dark compounds in case of emission from triplet excited states. These states have no influence on color of material and don't reveal absorption due to forbidden nature of 1S0->3Tn transitions. Now let's imagine the situation when 3Tn level lies in the transparent spectral window between two 1Sn and 1Sn-1 states. If the population of such state through intersystem crossing process takes place, in theory, emission may be observed.