Bragg diffraction and electron-gas heat capacity
Introduced| English |
|---|
| Bragg condition/bræɡ kənˈdɪʃn/ |
| Fermi energy/ˈfɜːmi ˈenədʒi/ |
A decision before an answer
- Rotating a crystal in an X-ray beam maps its atomic planes, because constructive reflection happens only at discrete angles.
- Your goal: Apply the Bragg condition to find plane spacings or diffraction angles.
Apply the Bragg condition
- Bragg reflection from parallel planes spaced d interferes constructively when 2d sinθ=nλ, with θ measured from the plane surface (the glancing angle), not from the normal. First order n=1 gives the smallest angle.
- For λ=0.154 nm and a peak at θ=20°, d=0.154/(2 sin20°)≈0.225 nm. If 2d<λ no order exists; the sine must stay at or below 1.
X-rays of wavelength 0.154 nm show a first-order Bragg peak at θ=20°. The plane spacing is about:
d=λ/(2 sinθ)=0.154/(2×0.342)≈0.225 nm.
Read the lattice constant
- In a cubic crystal the (hkl) plane spacing is d=a/sqrt(h²+k²+l²) with lattice constant a; the (110) planes give a/√2. A measured Bragg angle therefore measures the lattice constant.
- Systematic absences from the basis (for example body-centred lattices missing h+k+l odd) carry structure information beyond the spacing formula.
In the low-temperature metal heat capacity C=γT+βT³, the linear term γT comes from:
Only a T/T_F fraction of conduction electrons are excited, giving a linear-T electronic term; βT³ is the Debye lattice term.
Fill the electron sea
- The free-electron model packs N conduction electrons into states up to the Fermi wavevector k_F=(3π²n)^(1/3) for number density n, giving E_F=ℏ²k_F²/(2m).
- For n=8.5×10²⁸ /m³, k_F≈1.36×10¹⁰ /m, E_F≈7.05 eV and T_F=E_F/k_B≈8.2×10⁴ K. Since T_F far exceeds room temperature, only a fraction of order T/T_F of the electrons are thermally excited.
Cu Kα λ=0.154 nm with a first-order peak at θ=20° gives d≈0.225 nm; for cubic a=0.4 nm the (110) spacing is 0.283 nm. The free-electron count for n=8.5×10²⁸ /m³ gives E_F≈7.05 eV and T_F≈8.2×10⁴ K, so at 300 K only about 0.4% of electrons are thermally active. In C=γT+βT³, γT is electronic and βT³ is lattice.
For λ=0.10 nm and plane spacing d=0.25 nm, the first-order Bragg condition gives sinθ=____.
sinθ=λ/(2d)=0.10/0.50=0.2.
Separate the heat capacities
- Metals therefore have two low-temperature heat-capacity terms: an electronic term γT linear in T and a lattice (Debye) term βT³. Plotting C/T against T² gives intercept γ and slope β.
- At room temperature the lattice part approaches the classical Dulong–Petit value and dominates; the electronic term matters at low T, where the T³ lattice term collapses faster.
Measuring θ from the plane normal instead of the plane, or assigning the whole low-T heat capacity to electrons. The sine condition and the two-term decomposition are where these calculations fail.
Which answer fits this case?
Apply the Bragg condition to find plane spacings or diffraction angles
Doubling the X-ray wavelength halves the first-order Bragg angle for a given plane family.
The sine doubles, not the angle; and no order exists once λ/(2d) exceeds 1.
Keep the distinctions
- Bragg condition 布拉格条件 — Constructive reflection from lattice planes when 2d sinθ=nλ, with θ measured from the plane.
- Fermi energy 费米能 — The highest occupied electron energy at zero temperature in the free-electron model.
- Apply the Bragg condition to find plane spacings or diffraction angles.
- Estimate free-electron Fermi energy and Fermi temperature.
- Distinguish electronic and lattice heat-capacity contributions.
Match each term with its precise meaning in this lesson.
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
The order follows the stated process; check each stage before the next.