การกระจายของบรา็ก และความจุความร้อนของแก๊สอิเล็กตรอน
GRE Subject Test · GRE Physics หัวข้อ 45 5:29 การบรรยายภาษาอังกฤษ · คำบรรยายภาษาอังกฤษ + 中文 ลอยตัวบนภาพ
บท
Transcript
Rotating a crystal in an X-ray beam maps its atomic planes, because strong reflection happens only at certain angles.
在 X 射线束中转动晶体可以测出其原子平面，因为强反射只在特定角度出现。
This topic applies the Bragg condition, estimates the Fermi energy of the electrons in a metal, and separates the electronic and lattice heat capacities.
本主题应用布拉格条件，估算金属中电子的费米能，并区分电子热容和晶格热容。
Bragg reflection from parallel planes spaced d apart interferes constructively when 2 d sine theta equals n lambda, with theta measured from the plane surface, the glancing angle, not from the normal.
间距为 d 的平行晶面发生布拉格反射，当 2d sin θ 等于 n λ 时相长干涉，其中 θ 从晶面表面量起（掠射角），而不是从法线量起。
First order, n equal to 1, gives the smallest angle.
一级，即 n 等于 1，对应最小角度。
For lambda equal to 0.154 nanometres and a peak at theta equal to 20 degrees, d equals 0.154 over 2 sine 20 degrees, about 0.225 nanometres.
当 λ 等于 0.154 纳米、峰值位于 θ 等于 20 度时，d 等于 0.154 除以 2 sin 20 度，约为 0.225 纳米。
If 2 d is less than lambda, no order exists; the sine must stay at or below 1.
若 2d 小于 λ，则不存在任何级次；正弦值必须不大于 1。
In a cubic crystal, the hkl plane spacing is d equals a divided by the square root of h squared plus k squared plus l squared, where a is the lattice constant.
在立方晶体中，hkl晶面间距d等于a除以根号下h平方加k平方加l平方，其中a为晶格常数。
The 110 planes give a divided by the square root of 2.
110晶面对应a除以根号2。
A measured Bragg angle therefore measures the lattice constant.
因此，测得的布拉格角可用于测定晶格常数。
Systematic absences from the basis, for example body-centred lattices missing h plus k plus l odd, carry structure information beyond the spacing formula.
由基元引起的系统消光，例如体心晶格中h加k加l为奇数时缺峰，提供了超出间距公式的结构信息。
The free-electron model packs N conduction electrons into states up to the Fermi wavevector k sub F equals the cube root of 3 pi squared n, for number density n, giving E sub F equals h-bar squared k sub F squared divided by 2 m.
自由电子模型将N个传导电子填充至费米波矢k_F等于3π²n的立方根的状态，其中n为数密度，由此得到E_F等于h-bar平方乘以k_F平方除以2m。
For n equals 8.5 times 10 to the 28 per cubic metre, k sub F is approximately 1.36 times 10 to the 10 per metre, E sub F is approximately 7.05 electron volts, and T sub F equals E sub F divided by k sub B is approximately 8.2 times 10 to the 4 Kelvin.
当n等于8.5×10²⁸每立方米时，k_F约为1.36×10¹⁰每米，E_F约为7.05电子伏特，T_F等于E_F除以k_B约为8.2×10⁴开尔文。
Since T sub F far exceeds room temperature, only a fraction of order T over T sub F of the electrons are thermally excited.
由于T_F远高于室温，仅有数量级为T/T_F的电子被热激发。
Metals therefore have two low-temperature heat-capacity terms; an electronic term, gamma T, linear in T, and a lattice, or Debye, term, beta T cubed.
因此，金属在低温下有两个热容项：与 T 成正比的电子项 γT，以及晶格（德拜）项 β 乘以 T 的立方。
Plotting C over T against T squared gives intercept gamma and slope beta.
以 C 除以 T 对 T 的平方作图，截距为 γ，斜率为 β。
At room temperature the lattice part approaches the classical Dulong–Petit value and dominates; the electronic term matters at low T, where the T cubed lattice term collapses faster.
室温下晶格部分趋近经典的杜隆–珀蒂值并占主导；电子项在低温下才重要，因为此时 T 立方的晶格项下降得更快。
Measure theta from the crystal plane, not its normal.
θ 要从晶面量起，而不是从法线量起。
With 0.154 nanometre X-rays and a first-order peak at 20 degrees, d is 0.154 over 2 sine 20 degrees: 0.225 nanometres.
用 0.154 纳米的 X 射线，在 20 度处出现一级峰，则 d 等于 0.154 除以 2 sin 20°：0.225 纳米。
A metal at low temperature has an electronic term gamma T and a lattice term beta T cubed.
低温下金属的热容有电子项 γT 和晶格项 βT³。
Dividing by T gives C over T equals gamma plus beta T squared: a straight line against T squared, with intercept gamma and slope beta.
除以 T 得到 C/T 等于 γ 加 βT²：对 T² 作图是一条直线，截距为 γ，斜率为 β。
Original fictional molar heat capacity data give C divided by T equals 2.8 millijoules per mole kelvin squared at T equals 2 kelvin, and 5.2 in the same units at T equals 4 kelvin.
原始虚构的摩尔热容数据给出：在T等于2开尔文时，C除以T等于2.8毫焦每摩尔开尔文平方；在T等于4开尔文时，该值为5.2，单位相同。
Fit gamma and beta, then find the temperature at which electronic and lattice contributions are equal.
拟合gamma和beta，然后找出电子贡献与晶格贡献相等的温度。
Five marks.
五分。
Beta equals five point two minus two point eight, all over sixteen minus four, which is zero point two zero millijoules per mole kelvin to the fourth power.
β等于5.2减2.8，除以16减4，结果为0.20 mJ/(mol·K⁴)。
Gamma equals two point eight minus zero point twenty times four, which is two point zero millijoules per mole kelvin squared.
γ等于2.8减0.20乘4，结果为2.0 mJ/(mol·K²)。
Equality requires gamma T equals beta T cubed for T positive, hence T equals the square root of gamma divided by beta, which is the square root of ten kelvin, or three point one six kelvin.
等式要求γT等于βT³（T为正数），因此T等于√(γ/β)，即√10 K，或3.16 K。
Award transformed axes, slope, intercept, condition and temperature slash units.
评分标准包括变换坐标轴、斜率、截距、条件以及温度与单位。
Two constructed points cannot independently establish the model or experimental uncertainty.
两个构造点无法独立确立模型或实验不确定度。
Measuring θ from the plane normal instead of the plane, or assigning the whole low-T heat capacity to electrons.
从晶面法线而非晶面本身测量θ角，或将全部低温热容归因于电子。
The sine condition and the two-term decomposition are where these calculations fail.
正弦条件和两项分解正是这些计算失效之处。
Bragg condition: Constructive reflection from lattice planes when 2d sinθ=nλ, with θ measured from the plane.
布拉格条件：当2d sinθ = nλ时，来自晶面的相干反射，其中θ为相对于晶面的夹角。
Fermi energy: The highest occupied electron energy at zero temperature in the free-electron model.
费米能：自由电子模型中零温下被占据的最高电子能量。
Retrieve the meanings without reading the definitions, then change a parameter in the original transfer and test the limiting case.
不阅读定义，直接回忆其含义；随后改变原始传递中的一个参数，并检验极限情况。