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Mars rover Curiosity's ChemCam uses a pulsed diode-pumped solid-state laser (DPSS) at 1064 nm to excite material on the surfaces of rocks up to 7 meters away, and a large aperture telescope to collect emitted light for analysis.

From NASA JPL's MSL Science Corer page Chemistry & Camera (ChemCam):

LIBS Instrument

The LIBS instrument uses powerful laser pulses, focused on a small spot on target rock and soil samples within 7 m of the rover, to ablate atoms and ions in electronically excited states from which they decay, producing light-emitting plasma. The power density needed for LIBS is > 10 MW/mm^2, which is produced on a spot in the range of 0.3 to 0.6 mm diameter using focused, ~14 mJ laser pulses of 5 nanoseconds duration. The plasma light is collected by a 110 mm diameter telescope and focused onto the end of a fiber optic cable. The fiber carries the light to three dispersive spectrometers which record the spectra over a range of 240 - 850 nm at resolutions from 0.09 to 0.30 nm in 6144 channels. The spectra consist of emission lines of elements present in the samples. Typical rock and soil analyses yield detectable quantities of Na, Mg, Al, Si, Ca, K, Ti, Mn, Fe, H, C, O, Li, Sr, and Ba. Other elements often seen in soils and rocks on Earth include S, N, P, Be, Ni, Zr, Zn, Cu, Rb, and Cs. It is anticipated that 50-75 laser pulses will be required achieve the desired 10% accuracy for major elements at 7 m distance.

Wikipedia's Laser-induced breakdown spectroscopy says:

LIBS operates by focusing the laser onto a small area at the surface of the specimen; when the laser is discharged it ablates a very small amount of material, in the range of nanograms to picograms, which generates a plasma plume with temperatures in excess of 100,000 K. During data collection, typically after local thermodynamic equilibrium is established, plasma temperatures range from 5,000–20,000 K. At the high temperatures during the early plasma, the ablated material dissociates (breaks down) into excited ionic and atomic species. During this time, the plasma emits a continuum of radiation which does not contain any useful information about the species present, but within a very small timeframe the plasma expands at supersonic velocities and cools. At this point the characteristic atomic emission lines of the elements can be observed. The delay between the emission of continuum radiation and characteristic radiation is in the order of 10 μs, which is why it is necessary to temporally gate the detector.

Question: What is the process by which infrared 1064 nm (1.17 eV) photons can produce an ionized plasma that results in the emission of UV photons as short as 240 nm (5.17 eV)? how exactly does a beam of 1.17 eV photons initially produce a "plasma plume with temperatures in excess of 100,000 K" where $k_B T$ = 8.6 eV.


NASA JPL's MSL Science Corer diagram of ChemCam

NASA JPL's MSL Science Corer low quality example of emitted optical spectra from IR to UV, image sharpened

Images from NASA.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – andselisk
    Sep 17, 2020 at 18:12
  • $\begingroup$ I try to convince OP that the critical passage he made is "Question: What is the process by which infrared 1064 nm (1.17 eV) photons can produce an ionized plasma that results in the emission of UV photons as short as 240 nm (5.17 eV)? how exactly does a beam of 1.17 eV photons initially produce a "plasma plume with temperatures in excess of 100,000 K" where = 8.6 eV.". He seems to treat the laser as a black body. There is no upconversion of any photon. There is just heating of the sample through nir absorption. The heating might be difficult to model for shape/gradient , but it is heating. $\endgroup$
    – Alchimista
    Sep 19, 2020 at 10:20
  • $\begingroup$ @Alchimista No, I do not seem to do that at all! However your comment seems to suggest I don't understand the distinction. It's the exact opposite. The distinction is absolutely at the core of my question. Please double check my reply to your previous comment. The question sets up the difference between the two as part of the problem, and thermalization and implications of equipartition will be the nature of the answer. $\endgroup$
    – uhoh
    Sep 19, 2020 at 10:23

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