Atomic Absorption Spectroscopy Questions and Answers

Answer: A

The technique measures absorption by free, unexcited atoms in the vapour, which is why the sample must first be atomised. Molecules in solution give broad bands rather than sharp lines, ions have different energy levels and are a loss rather than the analyte, and excited atoms returning to ground emit rather than absorb. That the measurement needs atoms, not molecules or ions, is what makes the atomiser the heart of the instrument. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

Answer: B

An isolated atom has only electronic levels, with no vibrational or rotational ladders to broaden the transition, so the line is a few thousandths of a nanometre wide. Concentration and flame temperature broaden it slightly rather than narrowing it, and the source does not determine the atom's line. That the absorbing line is far narrower than any monochromator can isolate is the fact that forces the whole unusual design of the instrument. Most errors here come from carrying a result across from a neighbouring case where it does not hold.

Answer: C

Because the atomic line is a hundred times narrower than the narrowest practical monochromator band, a continuum source would have almost all its transmitted light unabsorbed and the measured absorbance would be tiny. Source intensity, detector response and flame absorption are not the reason. Using a source that emits exactly the line the atom absorbs sidesteps the resolution problem entirely, and it is why a separate lamp is needed for each element. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

Answer: D

The cathode is lined with the analyte metal, so sputtering and excitation produce that element's own emission lines - which is exactly what its atoms in the flame will absorb. A tungsten filament and a deuterium discharge are continuum sources, and mercury vapour would emit mercury's lines. A source built from the analyte itself guarantees a perfect match in wavelength, and the price is one lamp per element. Most errors here come from carrying a result across from a neighbouring case where it does not hold.

Answer: A

Ionised filler gas is accelerated into the cathode and knocks metal atoms into the vapour, where the discharge excites them to emit. Absorption, cooling and oxidation protection are incidental. The choice between neon and argon depends on which gives the cleaner spectrum near the analytical line for a given element, since the filler's own lines are potential interferences.

Answer: B

Running the lamp hard produces a dense cloud of ground-state atoms in front of the cathode, which absorb the centre of the emitted line and broaden it - self-absorption. A broadened source line no longer matches the narrow atomic absorption and sensitivity falls. Lamp life does suffer too, which is a real but secondary consideration. Whenever a source and an absorber are the same element, the source's own vapour is a hazard worth designing against. Framing the question as which quantity the change acts upon makes the answer follow without calculation.

Answer: C

Applying the Boltzmann distribution with a 2.1 electron volt gap:

\[ \begin{aligned} \frac{N^{*}}{N_{0}} &\approx e^{-2.1/kT} \\ &\approx 6\times10^{-5} \end{aligned} \]

which is six parts in a hundred thousand. Essentially every atom is in the ground state, which is why absorption by the ground state is so much more reliable than emission from the excited one - and why atomic absorption is less sensitive to flame temperature than atomic emission.

Answer: D

Since almost every atom is already in the ground state, a change in temperature moves that population hardly at all, while the tiny excited population that emission relies on changes by a large factor. Flame emission is chopped out rather than subtracted automatically, the lamp compensates for nothing, and absorption certainly depends on atom number. Measuring a quantity that is already near its maximum makes a method robust against the variable that determines it.

Answer: A

Air-acetylene at about 2300 degrees atomises most metals adequately and is cheap, safe and stable, which is why it is the default. The hotter nitrous oxide flame is reserved for refractory elements that air-acetylene cannot break down. Air-hydrogen has a low background but is cooler still, and oxygen-acetylene burns too fast to be safe on a burner of this kind. Choosing the coolest flame that does the job keeps ionisation and background low. The quantity that varies and the quantity that does not are worth naming before any substitution is attempted.

Answer: B

Aluminium, silicon, titanium and the rest form oxides too stable for an air-acetylene flame to break, so the hotter nitrous oxide flame is required. Readily ionised elements are made worse by the extra heat rather than better, and dissolved solids and wavelength are separate issues. A hotter flame solves one problem and creates another, which is why it is used only where the cooler one fails. A single measurement chosen to respond to the one asymmetry beats several that respond to shared behaviour.