GATE Spectroscopy — Beer–Lambert Numericals
The Beer–Lambert law is one line of algebra, which is precisely why GATE Chemistry can build so many different numericals on it. The questions rarely ask you to state the law; they ask you to convert transmittance to absorbance and back, to extract a molar absorptivity, to work backwards through a dilution, to solve a two-component mixture by simultaneous equations, or to say why a calibration plot curved. This guide works through each of those types with the arithmetic done in full.
The law and its variables
| Symbol | Meaning | Usual unit |
|---|---|---|
| A | Absorbance (optical density) | dimensionless |
| ε | Molar absorptivity (molar extinction coefficient) | L mol⁻¹ cm⁻¹ |
| c | Concentration | mol L⁻¹ |
| l | Path length of the cell | cm |
| T | Transmittance I/I₀ | fraction (or ×100 for %T) |
ε is a property of the substance at one wavelength, not of the sample, and it is the quantity that ties spectroscopy to structure. Fully allowed charge-transfer and π→π* bands run to 10⁴–10⁵ L mol⁻¹ cm⁻¹. Laporte-forbidden d–d transitions in centrosymmetric octahedral complexes are typically only ~1–100, which is why those solutions are pale; tetrahedral complexes, having no inversion centre, show intensities an order or two higher and are correspondingly deeply coloured.
Type 1 — absorbance ⇄ transmittance
(a) A solution has A = 0.30. What is its percentage transmittance?
T = 10−0.30 = 0.5012, so %T = 50.1%. (The useful mental anchor:
A = 0.3 is roughly "half the light through", A = 1 is exactly 10%, A = 2 is 1%.)
(b) A solution transmits 25.0% of the incident light. If ε = 1 200 L mol⁻¹ cm⁻¹
in a 1.00 cm cell, find c.
A = −log10(0.250) = 0.6021
c = A / (εl) = 0.6021 / (1 200 × 1.00) = 5.02 × 10⁻⁴ mol L⁻¹
Type 2 — path length scaling
Absorbance is linear in path length; transmittance is exponential in it. Handle these questions in absorbance and convert only at the end.
A solution transmits 60.0% in a 1.00 cm cell. What does it transmit in a 3.00 cm cell?
A₁ = −log10(0.600) = 0.2218
A₃ = 3 × 0.2218 = 0.6654
T₃ = 10−0.6654 = 0.2160 → 21.6%
Check the shortcut: T₃ = (T₁)³ = 0.600³ = 0.216. The two routes agree, which is a free verification you should always take.
Type 3 — molar absorptivity and dilution back-calculation
(a) A 2.50 × 10⁻⁵ M solution gives A = 0.850 in a 1.00 cm cell. Find ε.
ε = A/(cl) = 0.850 / (2.50 × 10⁻⁵ × 1.00) =
3.40 × 10⁴ L mol⁻¹ cm⁻¹ — an intensity that indicates a fully allowed
transition.
(b) A stock solution absorbs too strongly, so 5.00 mL is diluted to 50.00 mL. The diluted sample reads A = 0.640 in a 1.00 cm cell with ε = 8 000 L mol⁻¹ cm⁻¹. Find the stock concentration.
Diluted: c = 0.640 / (8 000 × 1.00) = 8.00 × 10⁻⁵ M
Dilution factor = 50.00 / 5.00 = 10.0
Stock = 8.00 × 10⁻⁵ × 10.0 = 8.00 × 10⁻⁴ M
The dilution factor multiplies, never divides, when going back to the stock. Sketching the direction ("stock is more concentrated") prevents the classic sign-of-thinking error.
Type 4 — two-component mixture
Absorbances of non-interacting species are additive at every wavelength. With two absorbing species you need measurements at two wavelengths, giving two linear equations.
In a 1.00 cm cell: at 400 nm, εX = 12 000 and εY = 2 000; at 500 nm, εX = 3 000 and εY = 9 000 (all L mol⁻¹ cm⁻¹). The mixture gives A(400) = 0.750 and A(500) = 0.560. Find both concentrations.
12 000 cX + 2 000 cY = 0.750 … (i)
3 000 cX + 9 000 cY = 0.560 … (ii)
From (i): cX = (0.750 − 2 000 cY) / 12 000. Substituting into (ii):
0.25(0.750 − 2 000 cY) + 9 000 cY = 0.560
0.1875 − 500 cY + 9 000 cY = 0.560
8 500 cY = 0.3725 → cY = 4.38 × 10⁻⁵ M
cX = (0.750 − 0.08765) / 12 000 = cX = 5.52 × 10⁻⁵ M
Verify in (ii): 3 000(5.520 × 10⁻⁵) + 9 000(4.382 × 10⁻⁵) = 0.1656 + 0.3944 = 0.5600 ✓
Why calibration plots curve — real deviations
GATE regularly asks for the cause, so keep the three categories separate.
| Category | Cause | Effect on the A-vs-c plot |
|---|---|---|
| Real (fundamental) | Above ~0.01 M, solute molecules are close enough to alter each other's ε; refractive index changes with c | Curvature at high concentration |
| Chemical | Analyte dissociates, associates, dimerises or reacts with solvent; acid–base equilibria shift with dilution | Curvature either way; fixed by buffering or choosing an isosbestic point |
| Instrumental | Polychromatic radiation (finite bandwidth) and stray light | Always negative deviation; worst at high A and on a steep part of the band |
Two practical consequences follow. First, measure on an absorbance maximum, where the band is flat, so a finite slit width samples nearly constant ε. Second, keep A in roughly the 0.2–0.8 window: at very low A the signal is buried in noise, and at very high A a small transmittance error becomes a large concentration error. For a constant uncertainty in T, the relative error in concentration is smallest near A ≈ 0.434 (T ≈ 36.8%) — a result worth quoting if the question asks for the optimum working range.
Common mistakes
- Using natural log. The decadic law uses log₁₀. If a question gives the Napierian form A = ε′cl with ln, the two ε values differ by a factor of ln 10 = 2.303.
- Feeding %T straight into the log. Convert 25% to the fraction 0.25 first; −log(25) is nonsense.
- Path length in millimetres. ε is quoted per centimetre. A 2 mm cell is l = 0.2 cm.
- Dividing by the dilution factor when working back to a stock. The stock must come out more concentrated than the measured sample.
- Adding transmittances. Absorbances are additive for a mixture; transmittances multiply.
- Blaming "deviation from Beer's law" for a chemical equilibrium. If the absorbing species itself is changing with dilution, the law is not failing — the composition is.
Summary of the working relations
| You are given | You want | Use |
|---|---|---|
| %T | A | A = 2 − log₁₀(%T) |
| A | T | T = 10−A |
| A, ε, l | c | c = A/(εl) |
| A, c, l | ε | ε = A/(cl) |
| A at l₁ | A at l₂ | A₂ = A₁ (l₂/l₁) |
| Calibration line | ε | slope of A vs c = εl |
| Two λ, two species | cX, cY | Solve the 2 × 2 linear system |
Do the numbers, then check them. The Beer–Lambert calculator takes any three of A, ε, c and l and returns the fourth, and converts between absorbance and percentage transmittance — exactly the steps where a misplaced power of ten turns correct reasoning into a wrong option.
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