브래그 회절 및 전자 기체 열용량
도입됨| English | 의미 |
|---|---|
| Bragg's law/bræɡz lɔː/ | The condition 2d sin theta = n lambda for constructive reflection from crystal planes. |
| Fermi energy/ˈfɜːmi ˈenədʒi/ | Fermi能量: Fermi能量 |
| lattice constant/ˈlætɪs ˈkɒnstənt/ | The side length of a crystal’s repeating unit cell. |
Planes reveal themselves at set angles
- Turn a crystal in an X-ray beam: strong reflections flash out only at particular angles, each tied to a family of atomic planes.
- GRE solid-state questions test Bragg's law 布拉格定律, cubic plane spacings, the free-electron Fermi energy 费米能 and low-temperature heat capacity.
Bragg's law
- $2d\sin\theta = n\lambda$, with $\theta$ the glancing angle from the plane, not the normal. $\sin\theta \le 1$ limits the orders.
- $\lambda = 0.154$ nm, first-order peak at $\theta = 20°$: $d = 0.154/(2\sin20°) \approx 0.225$ nm.
$\lambda = 0.154$ nm, $d = 0.30$ nm, first order. Bragg angle in degrees (one decimal place)?
난이도 2/5 · 140 코인 + 오늘 첫 완료 시 소액 보너스
Bragg's law $2d\sin\theta = \lambda$ gives $\sin\theta = 0.154/0.60 = 0.2567$, so $\theta = 14.9°$.
$d = 0.20$ nm, $\lambda = 0.154$ nm. Highest Bragg order observable?
난이도 2/5 · 140 코인 + 오늘 첫 완료 시 소액 보너스
Orders need $\sin\theta = n\lambda/2d \le 1$, so $n \le 2d/\lambda = 2.6$: the highest is 2.
In Bragg's law, $\theta$ is measured
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
Bragg angles are glancing angles measured from the lattice planes, not from their normal.
Cubic planes
- With lattice constant 晶格常数 $a$, $d_{hkl} = \dfrac{a}{\sqrt{h^2 + k^2 + l^2}}$: $(110)$ gives $a/\sqrt2$. For $a = 0.4$ nm, $d_{110} = 0.283$ nm.
- Systematic absences (body-centred: $h + k + l$ odd missing) reveal the basis.
Cubic lattice, $a = 0.50$ nm. Spacing of (111) planes in nm (three decimal places)?
난이도 2/5 · 140 코인 + 오늘 첫 완료 시 소액 보너스
For a cubic lattice $d = a/\sqrt{h^2 + k^2 + l^2} = 0.5/\sqrt3 = 0.289$ nm.
Free electrons
- $k_F = (3\pi^2n)^{1/3}$, $E_F = \dfrac{\hbar^2k_F^2}{2m}$. Copper, $n = 8.5\times10^{28}$/m³: $E_F \approx 7.0$ eV, $T_F \approx 8\times10^4$ K.
- Only a fraction $\sim T/T_F$ of electrons is thermally active — about 0.4% at room temperature.
Two heat-capacity terms
- Metals: $C = \gamma T + \beta T^3$ (electrons + lattice). Plot $C/T$ against $T^2$: intercept $\gamma$, slope $\beta$.
- Low $T$: the electronic term wins. Room temperature: the lattice dominates, near Dulong–Petit.
If $C/T = 0.7 + 0.05T^2$ (mJ mol⁻¹ K⁻²), then $\gamma = 0.7$ and $\beta = 0.05$; the two terms are equal where $0.05T^2 = 0.7$, $T \approx 3.7$ K.
Which term dominates?
Compare γT and βT³ contributions across temperatures.
At room temperature, the electronic term dominates the heat capacity of a metal.
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
At room temperature the lattice (phonon) term is near its classical value and far exceeds the small electronic term.
Recap
- $2d\sin\theta = n\lambda$, $\theta$ from the planes.
- $d_{hkl} = a/\sqrt{h^2 + k^2 + l^2}$.
- $E_F$ is several eV; $C = \gamma T + \beta T^3$ separates electrons from the lattice.
- Bragg's law: The condition 2d sin theta = n lambda for constructive reflection from crystal planes.
- lattice constant: The side length of a crystal’s repeating unit cell.