UV Visible Spectroscopy Questions and Answers

Answer: A

Inverting the logarithmic definition:

\[ T = 10^{-0.80} \approx 0.158 \]

giving about 16 per cent. Eighty per cent confuses the absorbance's numerical value with a percentage, which the logarithm makes impossible. Since each unit of absorbance is a tenfold reduction, a reading of 0.8 lets through rather less than a fifth of the light.

Answer: B

Anthracene's three fused rings give a much larger conjugated system, which both lowers the transition energy and raises the transition probability, so its bands are further to the red and far more intense. Benzene absorbs at shorter wavelength, not longer, and solvent and spin are not the reason. That extending conjugation moves a band on both axes at once - bathochromic and hyperchromic together - is the general pattern for these systems. A limit that survives making every external component ideal is an intrinsic one, and worth knowing for that reason.

Answer: C

The rules give an empirical estimate of where a diene or an enone will absorb, starting from a base value and adding increments for each substituent and ring feature. They say nothing about intensity, solvent transparency or band count. An empirical scheme built from many measured examples can be more useful than a calculation from first principles when the molecules are complicated, and these rules survived for decades on exactly that basis. The shared properties explain why these belong to one family, and the differences explain why more than one exists.

Answer: D

Adding each increment to the base value:

\[ 217 + 3 \times 5 = 232 \ \mathrm{nm} \]

giving 232 nanometres. The scheme is additive by construction, so the arithmetic is trivial and the skill lies in identifying which increments apply - an exocyclic double bond, a ring residue, an auxochrome each carrying its own value.

Answer: A

A deuterium discharge gives a smooth continuum from about 190 to 400 nanometres, where a tungsten filament's output has collapsed. Tungsten covers the visible and near infrared, a hollow cathode lamp emits an element's own lines for atomic absorption, and a globar is an infrared source. Since no one lamp covers the whole range, an instrument switches between two and the changeover wavelength is one of its specifications. Most errors here come from carrying a result across from a neighbouring case where it does not hold.

Answer: B

Deuterium falls away above about 350 nanometres and tungsten is useless below about 320, so a scan crossing that region must change lamps, and a small step in the baseline often appears there. Overheating, detector requirements and dispersion are not the reason. Recognising that a baseline discontinuity near 340 nanometres is a lamp changeover rather than a sample feature saves chasing an artefact that is built into the instrument. The shared properties explain why these belong to one family, and the differences explain why more than one exists.

Answer: C

A photomultiplier is fast, sensitive and covers the whole ultraviolet and visible range, which is what a scanning instrument needs. Thermocouples and pyroelectric detectors belong to the infrared, where photon detectors perform poorly, and lead sulphide serves the near infrared. Matching the detector to the photon energies available is what decides between them, and ultraviolet photons are energetic enough for a photoemissive device to work. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

Answer: D

Dispersing the light after the sample onto an array lets every wavelength be measured at once, so a full spectrum takes milliseconds rather than minutes - ideal for a chromatography detector or a fast reaction. Its resolution is usually worse, being set by the element spacing, it still disperses the light, and it exposes the sample to the full beam rather than less. Speed bought at the cost of resolution and sample exposure is the trade, and it suits some applications exactly.

Answer: A

Because dispersion happens after the sample, the whole output of the lamp passes through the solution rather than one narrow band at a time. The source is comparable, the measurement is far quicker, and the detector is nowhere near the sample. Reversing the order of two components to gain speed has moved a stress onto the sample, which is a consequence worth anticipating before a photolabile compound is put in one. The shared properties explain why these belong to one family, and the differences explain why more than one exists.

Answer: B

Accuracy is the agreement between the wavelength displayed and the wavelength actually delivered, checked against a holmium filter or a mercury line. Bandwidth is resolution, agreement between scans is repeatability, and the ultraviolet limit is the spectral range. Four specifications describing four different things, and quoting one where another is meant makes an instrument comparison meaningless.