Spectrophotometry Fundamentals Questions and Answers

Answer: C

A spectrum shows where the compound absorbs and how sharply, from which the analytical wavelength is chosen and the chromophores present may be inferred. Concentration follows only once the absorptivity is known, and neither stray light nor path length is displayed. Running a full spectrum before setting up a quantitative method is what identifies the best wavelength and warns of interfering absorption, and skipping it is how a method comes to be built on the wrong peak.

Answer: D

The spectrum shows where the analyte absorbs most strongly and most flatly, and whether anything else in the sample absorbs there too. Resolution, dark current and refractive index are instrument or solvent properties measured separately. Choosing an analytical wavelength without looking at the spectrum first is choosing blind, and it is the step most often skipped when a published method is adapted to a new matrix.

Answer: A

In solution an electronic transition is accompanied by a range of vibrational and rotational changes and is broadened further by solvent interaction, giving a band tens of nanometres wide. The slit adds instrumental broadening on top, but the underlying width is molecular. Neither concentration nor path length affects the shape, only the height. Distinguishing the natural width of a band from the width an instrument imposes is what decides whether a narrower slit would actually reveal anything more.

Answer: B

Once the instrument's bandwidth is a small fraction of the band being measured, further narrowing reveals no new structure and only loses light. Detector sensitivity, source output and absorbance range are separate limits. A resolution better than the thing being resolved buys nothing, which is a general principle worth applying before any instrument setting is pushed to its extreme.

Answer: C

A grating's angular dispersion is close to linear in wavelength, so the mechanical drive and the wavelength scale relate simply, whereas a prism's dispersion varies strongly across the range. A grating reflects rather than transmitting, still needs rotating, and produces stray light through its own imperfections and higher orders. A component chosen because it linearises a scale, rather than because it disperses more, is a recurring pattern in instrument design.

Answer: D

A grating diffracts each wavelength into several orders, so light of half the wavelength in the second order emerges at the same angle as the wanted first-order light. Detector harmonics, wall reflections and source lines are separate matters. Because a second-order overlap is indistinguishable from the wanted wavelength, an order-sorting filter is fitted to block it - which is one more component whose absence produces a puzzling error.

Answer: A

It passes the wanted order and blocks the shorter wavelengths that would arrive in a higher order at the same angle. Stray light, source switching and attenuation are other functions. Without it a reading in the visible could be contaminated by ultraviolet light the grating has folded onto the same path, and the error would look like a departure from Beer's law rather than an instrumental artefact.

Answer: B

Successive orders satisfy the grating equation at the same angle when:

\[ \begin{aligned} m\lambda &= \text{constant} \Rightarrow \lambda_{2} \\ &= \frac{600}{2} \\ &= 300 \ \mathrm{nm} \end{aligned} \]

giving 300 nanometres. Doubling rather than halving would place it outside the grating's range in that order. That the contaminating wavelength is always shorter, and so more energetic, is why an order-sorting filter is a long-pass one.

Answer: C

Infrared sources are weak and detectors noisy, so an instrument that measures every wavelength simultaneously rather than scanning them one at a time gains enormously in signal-to-noise for the same measurement time. Its resolution is set by the mirror travel, a cell is still needed, and the moving mirror is central to it. Where the limitation is a shortage of photons, collecting all of them at once is the decisive advantage, which is why the technique displaced dispersive infrared entirely.

Answer: D

Resolution is inversely proportional to the maximum optical path difference, so a longer mirror travel resolves finer structure. There is no dispersing slit, a single detector is used, and the source sets the range rather than the resolution. That resolution depends on a mechanical travel rather than on an optical slit is the structural difference from a dispersive instrument, and it is why the two are specified in quite different terms.