Difraksi Bragg dan kapasitas panas gas elektron
Diperkenalkan| English | Makna |
|---|---|
| Bragg condition/bræɡ kənˈdɪʃn/ | kondisi Bragg |
| Fermi energy/ˈfɜːmi ˈenədʒi/ | energi Fermi |
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:
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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:
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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=____.
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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.
Jawaban mana yang sesuai dengan kasus ini?
Terapkan kondisi Bragg untuk mencari jarak bidang atau sudut difraksi
Doubling the X-ray wavelength halves the first-order Bragg angle for a given plane family.
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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.
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Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
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The order follows the stated process; check each stage before the next.