Atomic Absorption Spectroscopy Questions and Answers

Answer: C

The technique measures free gaseous atoms, so the sample must be broken down to them by a flame or a furnace. Crystallisation, oxidation and neutralisation are sample preparation steps that may or may not be needed. That everything upstream of the atomiser exists only to deliver atoms into the light path is the organising idea of the whole instrument.

Answer: D

Nearly all determinations start from a solution, which a nebuliser converts to an aerosol or a pipette delivers into a furnace. Powders and pellets must be dissolved or digested first, and only mercury is handled as a vapour. That solids require digestion, with its own risks of loss and contamination, is often the largest source of error in a determination and has nothing to do with the instrument at all. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour.

Answer: A

The nebuliser draws solution up by the gas flow and shatters it into droplets, of which the finest reach the flame. It removes nothing, does no heating, and gas mixing happens in the spray chamber beyond it. Only a few per cent of the aspirated solution reaches the flame, which is the principal reason a flame is so much less sensitive than a furnace holding the whole sample.

Answer: B

Large droplets would not evaporate and atomise in the flame's brief transit, so the chamber's baffles remove them and only the fine fraction proceeds. Nothing is enlarged, cooled or added. Discarding most of the sample to ensure that what remains behaves reproducibly is a deliberate trade of sensitivity for precision, and it is why the drain must be watched for a steady flow.

Answer: C

Burner height is adjusted so the beam passes where free atoms are most numerous, which differs from element to element. Too low and atomisation is incomplete; too high and the atoms have oxidised or dispersed. The primary zone is turbulent and the outer region is where oxidation occurs. Optimising a physical position for each analyte, rather than fixing it once, is part of setting up a method properly. Naming the cost of a choice alongside its benefit is what turns a fact into a design decision.

Answer: D

A slot burner ten centimetres long gives the beam a correspondingly long path through the atoms, and absorbance is proportional to path length exactly as in solution photometry. Distance from the optics, fuel economy and detector width are secondary. That the Beer-Lambert law applies to a flame as readily as to a cuvette, with the burner slot as the cell, is the connection worth making.

Answer: A

The strongest resonance line gives the greatest sensitivity, since it connects the heavily populated ground state to an excited one with a large transition probability. Freedom from overlap matters and sometimes forces a weaker line to be chosen instead, and neither wavelength nor monochromator performance decides it. Choosing the transition from the state where nearly all the atoms are is what makes the technique sensitive in the first place. Working one numerical case through both cases side by side fixes the distinction better than any rule of thumb.

Answer: B

A less sensitive line extends the working range upward, so a concentrated sample can be read directly instead of being diluted. It worsens rather than improves the detection limit, does not remove the need for correction, and lamp current is unaffected. Deliberately giving up sensitivity to gain range is a trade worth making when dilution would introduce more error than the weaker line does. A limit that survives making every external component ideal is an intrinsic one, and worth knowing for that reason.

Answer: C

Using the standard conversion for photon energy:

\[ E = \frac{1240}{213} \approx 5.8 \ \mathrm{eV} \]

giving about 5.8 electron volts. A gap that large makes the excited state population utterly negligible at flame temperatures, which is why absorption is so much the better measurement for such an element and emission so nearly useless.

Answer: D

Applying the same conversion:

\[ E = \frac{1240}{589} \approx 2.1 \ \mathrm{eV} \]

giving about 2.1 electron volts, small enough that a measurable fraction of sodium atoms is thermally excited in a flame - which is precisely why sodium can be determined by flame emission photometry while zinc cannot. The size of the energy gap decides which of the two measurements is practical.