In Bragg’s law nλ = 2d sinθ, what does the quantity 2d sinθ physically represent?

The extra path travelled by the ray reflected from the lower of two adjacent planes

The distance between the source and the detector

The wavelength of the diffracted beam inside the crystal

The extra path travelled by the ray reflected from the lower of two adjacent planes

The projection of the unit cell edge on the beam direction

Crystal planes in metals are spaced by about 0.1 to 0.3 nm. Why is X-ray radiation, rather than visible light, used to study them by diffraction?

Its wavelength is of the order of the interplanar spacing, so that λ ≤ 2d can be met

It is the only radiation that can penetrate a metal to a depth of several centimetres

It carries an electric charge that is deflected by the lattice planes

Its photons have lower energy than visible photons, so they do not damage the lattice

A cubic powder pattern has its first four lines at sin²θ = 0.0450, 0.0600, 0.1200 and 0.1650. Which lattice fits?

Face-centred cubic, since the ratios are 3 : 4 : 8 : 11

Body-centred cubic, since the ratios are close to 1 : 2 : 3 : 4

Face-centred cubic, since the ratios are 3 : 4 : 8 : 11

Diamond cubic, since the ratios are 3 : 8 : 11 : 16

Simple cubic, since every line is allowed

KCl has the rock-salt structure (an FCC lattice with K⁺ and Cl⁻ in the basis), yet its powder pattern looks like a simple cubic lattice of half the true lattice parameter. What is the reason?

K⁺ and Cl⁻ are isoelectronic with nearly equal scattering factors, so lines with all-odd indices, whose F is 4(f₁ − f₂), almost vanish

The all-even lines have F = 4(f₁ − f₂), which cancels for equal ions

KCl actually crystallises as a simple cubic lattice at room temperature

The X-ray wavelength is automatically halved inside an ionic crystal

Which of the following statements about vibrational spectroscopy are correct?

P: A vibration is IR active only if it changes the molecular dipole moment.

Q: A vibration is Raman active only if it changes the molecular polarisability.

P: A vibration is IR active only if it changes the molecular dipole moment.

Q: A vibration is Raman active only if it changes the molecular polarisability.

R: The symmetric stretch of linear CO₂ is IR active.

S: The Raman shift in cm⁻¹ of a given band changes when the laser wavelength is changed.

Which statement about the fundamental vibrations of the linear, centrosymmetric CO₂ molecule is correct?

The symmetric stretch is Raman active and IR inactive; the asymmetric stretch is IR active and Raman inactive

The symmetric stretch is IR active because both C=O bonds change length

The symmetric stretch is Raman active and IR inactive; the asymmetric stretch is IR active and Raman inactive

The asymmetric stretch is active in both IR and Raman spectra

All three fundamental vibrations are active in both techniques

Which of the following statements about microscopes are correct?

P: At comparable magnification a scanning electron microscope has a much larger depth of field than an optical microscope.

Q: A transmission electron microscope needs a specimen that is electron-transparent, typically well under a micrometre thick.

P: At comparable magnification a scanning electron microscope has a much larger depth of field than an optical microscope.

Q: A transmission electron microscope needs a specimen that is electron-transparent, typically well under a micrometre thick.

R: Increasing the accelerating voltage increases the electron wavelength.

S: The resolution of an electron microscope equals its electron de Broglie wavelength (picometres).

In energy dispersive spectroscopy attached to an electron microscope, which quantity identifies the element present?

The energy of the emitted characteristic X-ray photons

The angle at which the electron beam is diffracted from the lattice planes

The brightness of the secondary electron image at the point

The width of the transmitted beam after the sample

The energy of the emitted characteristic X-ray photons

Structure and Composition: X-ray Diffraction, Spectroscopy, Microscopy and EDS

This chapter belongs to the Engineering Sciences (XE) paper. It answers the first paragraph of Section XE2.4 of XE2 Materials Science, one of the optional sections of which a candidate chooses two in the exam: “X-ray diffraction: Bragg’s Law, structure factor, indexing of cubic diffraction patterns; spectroscopic techniques: UV-Vis, IR and Raman; band-gap measurement; Microscopy (optical, scanning and transmission electron microscopy): wavelength range, resolution, depth of field; Composition analysis using energy dispersive spectroscopy.” It teaches Bragg’s law, the structure factor and the systematic absences of SC, BCC and FCC lattices, indexing from sin²θ ratios, what UV-Vis, IR and Raman each probe and their selection rules, the Tauc plot, the wavelength, resolution and depth of field of optical, scanning and transmission electron microscopes, and EDS. The numericals are Bragg angle and d-spacing, indexing ratios, lattice parameter, electron wavelength, Abbe resolution and band gap from an absorption edge.

Bragg’s law and what a diffraction pattern gives

Treat crystal planes of spacing d as partially reflecting mirrors. Two rays reflected from adjacent planes differ in path by 2d sinθ, where θ is the glancing angle between the beam and the planes. They reinforce when the path difference is a whole number of wavelengths: nλ = 2d sinθ (Bragg’s law). Because sinθ ≤ 1, a reflection exists only if λ ≤ 2d/n. That is why X-rays (λ about 0.05 to 0.25 nm, comparable to atomic spacings) diffract from crystals while visible light cannot. The detector sits at 2θ from the incident direction, so patterns are recorded as intensity against 2θ.

For a cubic crystal of lattice parameter a, the spacing of the (hkl) planes is d = a/√(h² + k² + l²). Combining, sin²θ = (λ²/4a²)(h² + k² + l²). The peak positions therefore fix the lattice type and a; the peak intensities depend on what atoms sit where (the structure factor); the peak widths carry crystallite size and strain information, which this section does not ask you to compute.

🧠 Worked example

A cubic metal has a = 0.405 nm. Find 2θ of the (111) reflection for λ = 0.154 nm, first order. d₁₁₁ = 0.405/√3 = 0.2338 nm. sinθ = λ/2d = 0.154/(2 × 0.2338) = 0.3293, so θ = 19.23° and 2θ = 38.45°. Check: the answer must be below 180° and sinθ below 1; a longer λ would push θ up and eventually forbid the reflection.

⚠️ Trap: θ versus 2θ

The instrument and the question usually quote 2θ, but Bragg’s law uses θ. Halve the peak position before taking the sine. Also, “second-order reflection of (100)” is the same as first-order (200): nλ = 2d sinθ with n = 2 equals λ = 2(d/2) sinθ.

Structure factor, systematic absences and indexing cubic patterns

The structure factor adds the waves scattered by every atom j in the unit cell, at fractional coordinates (uⱼ, vⱼ, wⱼ), with atomic scattering factor fⱼ: F(hkl) = Σ fⱼ exp[2πi(huⱼ + kvⱼ + lwⱼ)]. The intensity is proportional to |F|². If F = 0 the reflection is absent even though Bragg’s law is satisfied, because waves from atoms inside the cell cancel. For a Bravais lattice with one atom type the rules follow from the lattice points alone:

Selection rules for cubic lattices (one atom per lattice point)
Lattice	Basis points	F ≠ 0 when	First lines (h²+k²+l²)
SC	(0,0,0)	all hkl	1, 2, 3, 4, 5, 6, 8
BCC	(0,0,0), (½,½,½)	h + k + l even	2, 4, 6, 8, 10, 12
FCC	(0,0,0), (½,½,0), (½,0,½), (0,½,½)	h, k, l all odd or all even (unmixed)	3, 4, 8, 11, 12, 16

For BCC, F = f[1 + e^iπ(h+k+l)], which is 2f for even h + k + l and 0 for odd. For FCC, F = f[1 + e^iπ(h+k) + e^iπ(h+l) + e^iπ(k+l)], which is 4f for unmixed indices and 0 for mixed ones. The diamond cubic structure (FCC plus a second FCC shifted by ¼,¼,¼) keeps only unmixed hkl and also removes those all-even reflections with h + k + l = 4n ± 2, so its first lines are 3, 8, 11, 16. For a compound such as NaCl (FCC lattice with a two-ion basis), unmixed hkl survive; all-even lines have F = 4(f₁ + f₂) and all-odd lines have F = 4(f₁ − f₂), so odd lines are weak when the two ions scatter almost equally.

Indexing a powder pattern. Convert each peak 2θ to sin²θ, divide all values by the smallest, and compare with the allowed N = h² + k² + l² sequences. SC gives 1:2:3:4:5:6:8; BCC gives 1:2:3:4:5:6:7:8 after dividing the allowed N (2, 4, 6, 8, 10, 12, 14, 16) by 2; FCC gives 3:4:8:11:12, i.e. 1:1.33:2.67:3.67:4. A first ratio of 1.33 means FCC; absence of a 7 (N = 7 cannot be written as a sum of three squares) distinguishes SC from BCC at high angle. Once the lattice is known, a = λ√N/(2 sinθ) from any line.

⚠️ Trap: BCC versus SC ratios

BCC and SC both begin 1:2:3:4:5:6, so a six-line pattern cannot tell them apart. The seventh line separates them: BCC has it (N = 14, ratio 7) while SC lacks it because 7 is not a sum of three squares. Always test the first ratio for FCC (1.33) and for diamond (second line 2.67 times the first) before settling.

UV-Vis, IR and Raman spectroscopy

Spectroscopy measures how a material exchanges energy with light. UV-Vis (about 200 to 800 nm) probes electronic transitions: valence to conduction band in semiconductors, d–d and charge-transfer transitions in ions and oxides, π→π* in organics. Infrared absorption (mid-IR 4000 to 400 cm⁻¹, i.e. 2.5 to 25 µm) probes molecular and lattice vibrations. Raman scattering uses a monochromatic (usually visible) laser; a small fraction of the scattered light has lost (Stokes) or gained (anti-Stokes) one vibrational quantum, and the Raman shift (in cm⁻¹, measured from the laser line) equals the vibrational wavenumber.

The three techniques compared
Technique	Process	Selection rule	Typical use
UV-Vis	Absorption or reflectance, electronic	Allowed electronic transitions; direct or indirect gap	Band gap, colour, concentration
IR	Absorption, vibrational	Dipole moment must change during the vibration	Functional groups, bonds
Raman	Inelastic scattering, vibrational	Polarisability must change during the vibration	Phases, defects, carbon forms, strain

For a molecule with a centre of symmetry the rule of mutual exclusion holds: a vibration that is IR active is Raman inactive and vice versa. In linear CO₂ the symmetric stretch leaves the dipole at zero but changes the polarisability, so it is Raman active only; the asymmetric stretch and the bend change the dipole, so they are IR active only. At room temperature Stokes lines are stronger than anti-Stokes lines because the excited vibrational level is thinly populated (Boltzmann factor). A Raman shift does not depend on the laser wavelength; the absolute scattered wavelength does.

Band-gap measurement from absorption: the Tauc plot

A photon is absorbed across the gap only if hν ≥ E_g, so the absorption coefficient α rises sharply at the absorption edge. A quick estimate is E_g = hc/(e λ_edge); in convenient units E_g (eV) ≈ 1240/λ_edge (nm). For a film, α comes from the measured absorbance A and thickness t as α = 2.303 A/t; for powders, diffuse reflectance R is converted by the Kubelka–Munk function F(R) = (1 − R)²/(2R), which stands in for α.

The Tauc relation is (αhν)^1/r = A(hν − E_g), with r = ½ for an allowed direct transition and r = 2 for an allowed indirect transition. Plot (αhν)² against hν for a direct-gap material, or (αhν)^1/2 for an indirect one; extend the straight-line part to the hν axis and read E_g at the intercept. Choosing the wrong exponent gives a curved plot and a wrong intercept.

🧠 Worked example

A film’s absorption edge is at 500 nm. With h = 6.626 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m/s, e = 1.602 × 10⁻¹⁹ C: E = hc/λ = (6.626 × 10⁻³⁴ × 3.00 × 10⁸)/(500 × 10⁻⁹) = 3.976 × 10⁻¹⁹ J = 2.48 eV. The shortcut gives 1240/500 = 2.48 eV, the same. Check: visible photons (400 to 700 nm) carry 1.8 to 3.1 eV; a material that absorbs only below 500 nm looks yellow to orange.

⚠️ Trap: edge versus tail

Defect and impurity states cause a sloping absorption tail below the gap (Urbach tail). The band gap is the intercept of the straight Tauc region, not the wavelength where the first trace of absorption appears. A direct-gap material with a poor Tauc fit using r = ½ may actually be indirect: test both exponents.

Optical, scanning and transmission electron microscopy

Resolution is the smallest separation at which two points are distinguished. The Abbe limit is d = 0.61 λ/(n sinα) = 0.61 λ/NA, where NA is the numerical aperture of the objective. For visible light (400 to 700 nm) and NA near 1.4 (oil immersion) the limit is about 0.2 µm. Smaller λ and larger NA both help. Depth of field (the range of heights in focus) falls as NA rises, roughly as 1/NA² for an optical microscope, so high-magnification optical images of rough surfaces are in focus only in a thin slice.

Electrons have a de Broglie wavelength λ = h/p. After acceleration through V, eV = p²/2m, so λ = h/√(2m e V), or 1.226/√V nm with V in volts (neglecting relativity). At high voltages the relativistic correction matters: λ = h/√[2meV(1 + eV/2mc²)]. Typical relativistically corrected values are 12.2 pm at 10 kV (12.3 pm without the correction), 3.70 pm at 100 kV and 2.51 pm at 200 kV, about 10⁵ times shorter than visible light. Lens aberrations, not λ, limit electron microscopes, so the NA is kept tiny (semi-angle of order 10⁻² rad), but the resolution still reaches the nanometre or better.

Three microscopes compared
Feature	Optical	SEM	TEM
Probe	Visible light, 400 to 700 nm	Focused electron beam, about 1 to 30 kV, scanned over the surface	Electron beam, typically 100 to 300 kV, through the sample
Signal	Reflected or transmitted light	Secondary electrons (topography), backscattered electrons (atomic-number contrast), X-rays	Transmitted and diffracted electrons: image and diffraction pattern
Resolution	About 0.2 µm	Nanometre scale, typically 1 to 10 nm	Sub-nanometre; atomic columns resolved
Depth of field	Smallest	Large (narrow beam convergence)	Thin specimen, not a constraint
Sample	Polished and etched	Bulk; conducting or coated; vacuum	Electron-transparent thin foil, roughly under 100 nm

Composition analysis by energy dispersive spectroscopy

When the electron beam of an SEM or TEM ejects an inner-shell electron, an outer electron drops in and the atom emits a characteristic X-ray whose energy equals the difference of the two shell energies. That energy is fixed by the atomic number Z (Moseley: √ν rises linearly with Z), so it names the element; the number of counts at that energy measures how much is there. An energy-dispersive detector (a silicon-based semiconductor detector) records every photon and sorts it by energy, giving the whole spectrum at once with an energy resolution of order 130 eV. Wavelength-dispersive spectrometers (Bragg diffraction from a crystal) resolve far better but scan slowly.

Qualitative: peaks labelled by element and line (Kα, Kβ, Lα). Overlaps occur when lines lie within the detector resolution, e.g. neighbouring L and K lines of different elements.
Quantitative: peak intensities are converted to weight or atomic percent with matrix corrections (atomic number, absorption, fluorescence); accuracy is best for flat, polished samples of heavier elements.
Limits: hydrogen and helium give no characteristic X-rays; the lightest detectable elements (Li to O) are poorly measured because their low-energy photons are absorbed in the sample and detector window; the analysed volume is a micrometre-scale interaction volume in an SEM, so spatial resolution is much worse than the image resolution.

🎯 Why EDS differs from XRD

XRD identifies phases (which crystal structure) from the long-range order; EDS identifies elements regardless of structure. Two polymorphs of the same compound give different XRD patterns but identical EDS spectra.

Key takeaways

Bragg’s law nλ = 2d sinθ with d = a/√(h²+k²+l²) for cubic crystals; the detector angle is 2θ, so halve it before using the law.
Allowed reflections: SC all; BCC h+k+l even; FCC h, k, l unmixed. Powder sin²θ ratios: BCC 1:2:3:4:5:6:7:8, FCC 3:4:8:11:12:16, diamond 3:8:11:16.
IR needs a dipole change, Raman a polarisability change; centrosymmetric molecules obey mutual exclusion; Raman shift is independent of the laser wavelength.
Band gap: E_g (eV) ≈ 1240/λ_edge (nm); Tauc plot (αhν)² for direct and (αhν)^1/2 for indirect transitions, intercept on the hν axis.
Resolution d = 0.61λ/NA; electron λ = 1.226/√V nm (non-relativistic); SEM has the largest depth of field, TEM needs a thin foil and gives sub-nanometre resolution.
EDS reads characteristic X-ray energies to name elements and counts to quantify them; H and He are invisible and light elements are poorly measured.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

In Bragg’s law nλ = 2d sinθ, what does the quantity 2d sinθ physically represent?
1. The distance between the source and the detector
2. The wavelength of the diffracted beam inside the crystal
3. The extra path travelled by the ray reflected from the lower of two adjacent planes
4. The projection of the unit cell edge on the beam direction
Show answer
Answer: C — The extra path travelled by the ray reflected from the lower of two adjacent planes
The two rays reflected from adjacent planes of spacing d, with glancing angle θ, differ in path by 2d sinθ. Constructive interference needs this to be a whole number n of wavelengths. The wavelength itself is a property of the radiation, not of this path.
Crystal planes in metals are spaced by about 0.1 to 0.3 nm. Why is X-ray radiation, rather than visible light, used to study them by diffraction?
1. Its wavelength is of the order of the interplanar spacing, so that λ ≤ 2d can be met
2. It is the only radiation that can penetrate a metal to a depth of several centimetres
3. It carries an electric charge that is deflected by the lattice planes
4. Its photons have lower energy than visible photons, so they do not damage the lattice
Show answer
Answer: A — Its wavelength is of the order of the interplanar spacing, so that λ ≤ 2d can be met
Bragg’s law needs sinθ = nλ/2d ≤ 1, so λ must not exceed 2d. Visible wavelengths (400 to 700 nm) are thousands of times larger than 2d, so no Bragg reflection exists. X-rays of about 0.05 to 0.25 nm satisfy the condition. X-ray penetration into metals is only micrometres to tens of micrometres, X-rays are uncharged, and their photons are far more energetic than visible ones, so the other reasons are false.
For a monatomic BCC crystal with one atom per lattice point of the conventional cell, which of the following planes gives no first-order reflection?
1. (110)
2. (200)
3. (211)
4. (100)
Show answer
Answer: D — (100)
For BCC the structure factor is f[1 + e^iπ(h+k+l)], which vanishes when h + k + l is odd. (100) has sum 1, so F = 0. The planes (110), (200) and (211) have sums 2, 2 and 4 and all reflect, with F = 2f. The (100) reflection of a BCC crystal appears only as the (200) line of the half-spacing planes.
A cubic powder pattern has its first four lines at sin²θ = 0.0450, 0.0600, 0.1200 and 0.1650. Which lattice fits?
1. Body-centred cubic, since the ratios are close to 1 : 2 : 3 : 4
2. Face-centred cubic, since the ratios are 3 : 4 : 8 : 11
3. Diamond cubic, since the ratios are 3 : 8 : 11 : 16
4. Simple cubic, since every line is allowed
Show answer
Answer: B — Face-centred cubic, since the ratios are 3 : 4 : 8 : 11
Dividing by 0.0150 gives 3, 4, 8 and 11, the allowed N = h² + k² + l² values of FCC: (111), (200), (220), (311). Diamond cubic shares 3, 8, 11 but has no line at 4 (the (200) is extinct) and continues 16. BCC would give 2, 4, 6, 8 and SC 1, 2, 3, 4; neither fits the ratios 1 : 1.33 : 2.67 : 3.67.
KCl has the rock-salt structure (an FCC lattice with K⁺ and Cl⁻ in the basis), yet its powder pattern looks like a simple cubic lattice of half the true lattice parameter. What is the reason?
1. K⁺ and Cl⁻ are isoelectronic with nearly equal scattering factors, so lines with all-odd indices, whose F is 4(f₁ − f₂), almost vanish
2. The all-even lines have F = 4(f₁ − f₂), which cancels for equal ions
3. KCl actually crystallises as a simple cubic lattice at room temperature
4. The X-ray wavelength is automatically halved inside an ionic crystal
Show answer
Answer: A — K⁺ and Cl⁻ are isoelectronic with nearly equal scattering factors, so lines with all-odd indices, whose F is 4(f₁ − f₂), almost vanish
In the rock-salt structure unmixed hkl survive; all-even lines have F = 4(f₁ + f₂) and all-odd lines have F = 4(f₁ − f₂). K⁺ and Cl⁻ each have 18 electrons, so f₁ ≈ f₂ and the all-odd lines are nearly extinct. The surviving all-even lines index as a simple cubic lattice of spacing a/2. The all-even F is a sum, not a difference, so it does not cancel.
Which of the following statements about vibrational spectroscopy are correct?
1. P: A vibration is IR active only if it changes the molecular dipole moment.
2. Q: A vibration is Raman active only if it changes the molecular polarisability.
3. R: The symmetric stretch of linear CO₂ is IR active.
4. S: The Raman shift in cm⁻¹ of a given band changes when the laser wavelength is changed.
Show answer
Answer: A — P: A vibration is IR active only if it changes the molecular dipole moment.; B — Q: A vibration is Raman active only if it changes the molecular polarisability.
P and Q are the two selection rules. The symmetric stretch of CO₂ leaves the dipole at zero throughout, so R is false (it is Raman active). The Raman shift is the energy difference between laser and scattered photon, i.e. the vibrational quantum, and does not depend on the laser, so S is false; only the absolute wavelength of the scattered line moves.
A thin film is known to have an allowed direct band gap. Which plot, extrapolated to zero, gives the band gap from its UV-Vis absorption data? (α is the absorption coefficient, hν the photon energy.)
1. (αhν)^1/2 against hν
2. α against λ
3. (αhν)² against hν
4. ln α against 1/hν
Show answer
Answer: C — (αhν)² against hν
The Tauc relation (αhν)^1/r = A(hν − E_g) has r = ½ for allowed direct transitions, so (αhν)² is linear in hν above the edge and its intercept on the hν axis is E_g. The exponent ½ in the (αhν)^1/2 plot belongs to indirect transitions (r = 2). Plotting α against λ gives only an edge position, not a linear extrapolation.
Which statement about the fundamental vibrations of the linear, centrosymmetric CO₂ molecule is correct?
1. The symmetric stretch is IR active because both C=O bonds change length
2. The symmetric stretch is Raman active and IR inactive; the asymmetric stretch is IR active and Raman inactive
3. The asymmetric stretch is active in both IR and Raman spectra
4. All three fundamental vibrations are active in both techniques
Show answer
Answer: B — The symmetric stretch is Raman active and IR inactive; the asymmetric stretch is IR active and Raman inactive
For a molecule with a centre of inversion, vibrations that are symmetric about the centre change polarisability but not dipole (Raman only), and antisymmetric ones change the dipole but not polarisability (IR only): mutual exclusion. Bond lengths do change in the symmetric stretch, but the two induced dipole changes cancel, so the net dipole stays zero.
Which of the following statements about microscopes are correct?
1. P: At comparable magnification a scanning electron microscope has a much larger depth of field than an optical microscope.
2. Q: A transmission electron microscope needs a specimen that is electron-transparent, typically well under a micrometre thick.
3. R: Increasing the accelerating voltage increases the electron wavelength.
4. S: The resolution of an electron microscope equals its electron de Broglie wavelength (picometres).
Show answer
Answer: A — P: At comparable magnification a scanning electron microscope has a much larger depth of field than an optical microscope.; B — Q: A transmission electron microscope needs a specimen that is electron-transparent, typically well under a micrometre thick.
SEM works with a very narrow convergence angle, which gives a large depth of field, so P is true. A TEM beam must pass through the sample, so Q is true. Wavelength is h/√(2meV) and falls as V rises, so R is false. Lens aberrations, not wavelength, limit electron microscope resolution to the nanometre or sub-nanometre range, far above the picometre wavelength, so S is false.
For monatomic cubic crystals with one atom per lattice point of the conventional cell, which reflections are allowed?
1. P: (200) in BCC
2. Q: (200) in FCC
3. R: (110) in FCC
4. S: (100) in simple cubic
Show answer
Answer: A — P: (200) in BCC; B — Q: (200) in FCC; D — S: (100) in simple cubic
(200) has h + k + l = 2, even, so BCC allows it; its indices (2, 0, 0) are all even, so FCC allows it; simple cubic allows every hkl. (110) has mixed parity (odd, odd, even are not all same), so FCC forbids it; this is why (110) is the strong first line of BCC but absent in FCC.
In energy dispersive spectroscopy attached to an electron microscope, which quantity identifies the element present?
1. The angle at which the electron beam is diffracted from the lattice planes
2. The brightness of the secondary electron image at the point
3. The width of the transmitted beam after the sample
4. The energy of the emitted characteristic X-ray photons
Show answer
Answer: D — The energy of the emitted characteristic X-ray photons
An inner-shell vacancy filled by an outer electron emits an X-ray photon whose energy is the difference of the shell energies and depends on atomic number, so the photon energy names the element. Diffraction angles give crystal structure, and image brightness reflects topography or mean atomic number, not a specific element.
Which of the following elements cannot emit a characteristic X-ray line in an EDS experiment, because after its only electron is ejected no electron remains to fill the vacancy?
1. Carbon
2. Copper
3. Hydrogen
4. Aluminium
Show answer
Answer: C — Hydrogen
Hydrogen has a single 1s electron; if the beam removes it, nothing can drop into the vacancy, so no characteristic X-ray is emitted (helium likewise has no outer-shell transition). Carbon, aluminium and copper all have outer electrons that fill K or L vacancies, although carbon’s low-energy line is hard to measure.
A cubic crystal has lattice parameter a = 0.405 nm. For X-rays of wavelength λ = 0.154 nm, find the diffraction angle 2θ (in degrees) of the first-order (200) reflection. Give the answer to one decimal place.
Numerical answer — type the value.
Show answer
Answer: 44.7
d₂₀₀ = a/√(4) = 0.405/2 = 0.2025 nm. sinθ = λ/2d = 0.154/(2 × 0.2025) = 0.3802, so θ = 22.35° and 2θ = 44.7°. Python check: 2·asin(0.154/0.405) = 44.698°.
A powder pattern is known to come from an FCC metal. Its first (lowest-angle) line has sin²θ = 0.0600. Find sin²θ of the fifth line. Give the answer to two decimal places.
Numerical answer — type the value.
Show answer
Answer: 0.24
FCC lines have N = 3, 4, 8, 11, 12, … for (111), (200), (220), (311), (222). The fifth line is (222), N = 12. sin²θ ∝ N, so sin²θ = 0.0600 × 12/3 = 0.2400. A tempting error is to use N = 5 (the fifth integer) or to include forbidden mixed indices such as (110).
Electrons are accelerated from rest through 10 kV in a scanning electron microscope. Ignoring relativistic effects, find their de Broglie wavelength in picometres. Use h = 6.626 × 10⁻³⁴ J s, m = 9.109 × 10⁻³¹ kg, e = 1.602 × 10⁻¹⁹ C. Give the answer to one decimal place.
Numerical answer — type the value.
Show answer
Answer: 12.3
λ = h/√(2meV). 2meV = 2 × 9.109e-31 × 1.602e-19 × 1e4 = 2.9186e-45. Square root = 5.4024e-23. λ = 6.626e-34/5.4024e-23 = 1.2265e-11 m = 12.3 pm. Cross-check: 1.226/√10000 nm = 0.01226 nm.
An optical microscope uses light of wavelength 550 nm with an oil-immersion objective of numerical aperture 1.40. Using the Abbe criterion d = 0.61 λ/NA, find the smallest resolvable separation in nanometres. Give the answer as an integer.
Numerical answer — type the value.
Show answer
Answer: 240
d = 0.61 × 550/1.40 = 335.5/1.40 = 239.6 nm, which rounds to 240 nm, about 0.24 µm, the usual optical limit. Using the wavelength in metres without converting would only change the unit, not the ratio.
The absorption edge of a semiconductor film lies at 400 nm. Estimate its band gap in eV using E = hc/λ with h = 6.626 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m/s and 1 eV = 1.602 × 10⁻¹⁹ J. Give the answer to two decimal places.
Numerical answer — type the value.
Show answer
Answer: 3.10
E = (6.626e-34 × 3.00e8)/(400e-9) = 1.9878e-25/4.00e-7 = 4.970e-19 J. Dividing by 1.602e-19 J/eV gives 3.102 eV, so 3.10 eV. Shortcut: 1240/400 = 3.10 eV, consistent.