Some Properties of Atoms — Question 10

Section 40-1 · Physics pages 6590–6591

Question 10 A finite-size form factor can resolve a uniformly charged sphere For a supplied scattering model of a uniform sphere of radius R, the normalized amplitude is F(q) = 3[sin(qR) −qR cos(qR)]/(qR)3. The wave-number transfer q has units of inverse length. Assume qR ≪1. Z R q Schematic; labels, not drawing lengths, specify the data. (a) Expand F through order (qR)2. (b) Find the leading intensity reduction 1 −|F|2 and infer R if it is 0.020 at q = 2.00 × 1010 m−1.
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Question 10 - Solution A finite-size form factor can resolve a uniformly charged sphere Z R q (a) Method and calculation x = qR : sin x −x cos x = x3/3 −x5/30 + O(x7). F(q) = 1 −(qR)2/10 + O((qR)4) . (b) Interpretation and further calculation |F|2 = 1 −(qR)2/5 + O((qR)4) ⇒1 −|F|2 ≃(qR)2/5. R ≃ √ 5(.020)/q = 1.58 × 10−11 m . Physical check The inferred radius is a leading-order model estimate, with qR about 0.316. The charge profile is supplied; real atomic electron densities are not uniform spheres.

Question 10

A finite-size form factor can resolve a uniformly charged sphere

For a supplied scattering model of a uniform sphere of radius R, the normalized amplitude is F(q)=3[sin⁡(qR)−qRcos⁡(qR)]/(qR)3F(q)=3[\sin(qR)-qR\cos(qR)]/(qR)^3. The wave-number transfer q has units of inverse length. Assume qR≪1qR\ll1.

Diagram shown in the original SVG worksheet.

  1. Expand F through order (qR)2(qR)^2.

  2. Find the leading intensity reduction 1−|F|21-|F|^2 and infer R if it is 0.020 at q=2.00×1010m−1q=2.00\times10^{10}\,\mathrm{m^{-1}}.

Question 10 - Solution

A finite-size form factor can resolve a uniformly charged sphere

Diagram shown in the original SVG worksheet.

(a) Method and calculation

x=qR:sin⁡x−xcos⁡x=x3/3−x5/30+O(x7).x=qR:\quad\sin x-x\cos x=x^3/3-x^5/30+O(x^7).F(q)=1−(qR)2/10+O((qR)4).\boxed{F(q)=1-(qR)^2/10+O((qR)^4)}.

(b) Interpretation and further calculation

|F|2=1−(qR)2/5+O((qR)4)⇒1−|F|2≃(qR)2/5.|F|^2=1-(qR)^2/5+O((qR)^4)\Rightarrow1-|F|^2\simeq(qR)^2/5.R≃5(.020)/q=1.58×10−11m.\boxed{R\simeq\sqrt{5(.020)}/q=1.58\times10^{-11}\,\mathrm m}.

Physical check

The inferred radius is a leading-order model estimate, with qR about 0.316. The charge profile is supplied; real atomic electron densities are not uniform spheres.