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Spectrophotometry Fundamentals Questions and Answers

Answer: A

In a diode array the sample is illuminated with the full beam and the dispersion happens after it, so every wavelength lands on its own detector element simultaneously. Nothing is scanned, the slit is not broadened, and no computation reconstructs it. Reversing the order of sample and monochromator is what makes the whole spectrum available in milliseconds, at the cost of exposing the sample to the full beam - which matters for photosensitive materials. The quantity that varies and the quantity that does not are worth naming before any substitution is attempted.

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42.

Consider the following statements about the Beer-Lambert law.

1. Absorbance is proportional to concentration

2. Absorbance is proportional to path length

3. Transmittance is proportional to concentration

Which of the statements given above is or are correct?

Answer: B

The first two are correct and the third is not: transmittance falls exponentially with concentration, which is precisely why absorbance is defined as its negative logarithm. Confusing the two is the commonest error in this topic and it produces a calibration that curves when it should be straight. A quantity defined specifically to linearise a relationship should not be interchanged with the one it replaced. Reading the definition literally, rather than recalling a formula, is the reliable route through questions of this shape.

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43.

Consider the following pairs of instrument part and its function.

Monochromator - Selecting a narrow wavelength band

Sample cell - Defining the path length

Detector - Converting light into a signal

Blank - Correcting for stray light

How many of the pairs given above are correctly matched?

Answer: C

The first three are correctly paired and the fourth is not, so three match. A blank corrects for absorbance by the solvent, cell and reagents; stray light is an instrument defect that a blank does not remove, since it affects the blank and the sample differently. Attributing a correction to the wrong mechanism leads to trusting readings that the correction never actually fixed.

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44.

Match each defect with the symptom it produces.

List I

A. Stray light

B. Mismatched cells

C. Wrong blank

List II

P. Curvature at high absorbance

Q. A constant offset between readings

R. A calibration line missing the origin

Select the correct match.

Answer: D

The pairing is A-P, B-Q, C-R. Each defect leaves its own signature - a bend, a shift between cells, or an intercept - and reading the signature backwards names the cause. That three quite different faults produce three distinguishable symptoms is what makes a calibration curve a diagnostic tool rather than merely a conversion table.

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45.

The negative logarithm of transmittance is called ______.

Answer: A

This is absorbance, sometimes called optical density in older writing. Extinction is an older term for the same idea and for the absorptivity coefficient, which is part of why it fell out of use; turbidity describes scattering by suspended matter; and reflectance is the fraction returned from a surface. The quantity is defined logarithmically for one reason alone, which is that it is then proportional to concentration. The quantity that varies and the quantity that does not are worth naming before any substitution is attempted.

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46.

Which of the following is not a term in the Beer-Lambert law?

Slit width

Molar absorptivity

Path length

Concentration

Answer: B

Slit width is an instrument setting rather than a term in the law, though it affects how well the law is obeyed by determining the bandwidth admitted. The other three multiply together to give the absorbance. Distinguishing the quantities a law relates from the instrumental conditions under which it holds is what separates a correct calculation from a correct measurement.

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Answer: C

Wavenumber is one over the wavelength, conventionally in reciprocal centimetres, so it is directly proportional to photon energy. Dividing by the speed of light gives a time, there is no square relation, and micrometres are just another length unit. Infrared work uses wavenumbers precisely because a scale linear in energy makes vibrational transitions evenly spaced, while ultraviolet and visible work keeps nanometres. Checking the units of each quantity catches a wrong formula before the arithmetic begins.

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48.

A wavelength of \( 2000 \ \mathrm{nm} \) corresponds to a wavenumber of

Answer: D

Converting through centimetres:

\[ \begin{aligned} \tilde{\nu} &= \frac{10^{7}}{\lambda_{\mathrm{nm}}} \\ &= \frac{10^{7}}{2000} \\ &= 5000 \ \mathrm{cm^{-1}} \end{aligned} \]

giving 5000 per centimetre. The factor of ten to the seventh converts nanometres to centimetres, and remembering it as a single constant is quicker than working through the units each time.

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49.

A band at \( 4000 \ \mathrm{cm^{-1}} \) lies at a wavelength of

Answer: A

Inverting the same relation:

\[ \lambda = \frac{10^{7}}{4000} = 2500 \ \mathrm{nm} \]

giving 2500 nanometres, in the near infrared. Because the two scales are reciprocal, a band that looks narrow on one looks wide on the other, which is worth bearing in mind when comparing spectra plotted in different units.

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Answer: B

Photon energy is Planck's constant times frequency, and frequency is the speed of light over wavelength, so energy goes as one over wavelength - which is exactly what wavenumber is. The counting description is true but is not the reason, units are irrelevant, and energy is linear in frequency rather than quadratic. A unit chosen so that a scale becomes linear in the quantity of interest is worth adopting, and it is why infrared spectra are plotted the way they are.

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