UV Visible Spectroscopy Questions and Answers

Answer: D

If something else in the sample absorbs at the analyte's peak, moving to a shoulder where the interferent is transparent can be worth the loss in sensitivity. A large absorptivity is an advantage rather than a problem, and solvent transparency and a flat peak are both reasons to stay. Sacrificing the ideal operating point to escape an interference is a routine compromise, and recognising when it is worth making is what distinguishes a method developer from a technician.

Answer: A

Using the convenient conversion for photon energy:

\[ \begin{aligned} E &= \frac{1240}{\lambda_{\mathrm{nm}}} \\ &= \frac{1240}{254} \\ &\approx 4.9 \ \mathrm{eV} \end{aligned} \]

giving about 4.9 electron volts, comfortably enough to promote an electron between molecular orbitals. The constant of 1240 electron-volt nanometres is worth memorising, since it converts between the two most commonly used units in one division.

Answer: B

Applying the same conversion:

\[ E = \frac{1240}{700} \approx 1.8 \ \mathrm{eV} \]

giving about 1.8 electron volts, against 3.1 at the violet end - so a visible photon's energy varies by less than a factor of two across the whole range the eye can see. That the visible band is barely one octave wide, while the ear covers ten, is a comparison worth carrying.

Answer: C

From 400 to 700 nanometres is a factor of 1.75, which is less than the doubling an octave requires. Eye sensitivity, atmospheric windows and the solar spectrum all help explain why the range sits where it does but not why it is narrow in this sense. Comparing a range in octaves rather than in units makes it commensurable with hearing and with radio bands, and the comparison is striking.

Answer: D

If the instrument's response time is comparable with the time it spends on each wavelength, the recorded band lags behind the true one and appears shifted, broadened and flattened. Baseline, absolute readings and the wavelength scale are not systematically affected. A measurement made faster than the apparatus can respond will always distort in the direction of the scan, and comparing a forward and a reverse scan is the quickest way to reveal it.

Answer: A

Either a slower scan or a reverse scan reveals it, and scanning more slowly is the direct test: if the spectrum is unchanged the original rate was adequate. A reverse scan is an equally good diagnostic, since the distortion reverses with it. Slit width and solvent change other things. Testing whether a result depends on a setting that ought not to affect it is the general way to establish that a measurement is trustworthy.