Some Properties of Atoms — Question 8

Section 40-1 · Physics pages 6586–6587

Question 8 A displaced electron cloud gives a simple atomic polarizability In a supplied classical response model, an electron cloud of charge −e is displaced a distance x opposite a uniform electric field of magnitude E. A restoring force of magnitude kx acts toward the nucleus. Define the dipole magnitude p = ex and p = αE. E x Schematic; labels, not drawing lengths, specify the data. (a) Derive the equilibrium displacement and polarizability. (b) Include the restoring energy and field coupling to find the minimum total energy.
Show solutionHide solution
Question 8 - Solution A displaced electron cloud gives a simple atomic polarizability E x (a) Method and calculation kx = eE ⇒x = eE/k . p = ex = (e2/k)E ⇒α = e2/k . (b) Interpretation and further calculation U(x) = 1 2kx2 −eEx, U′(x) = kx −eE = 0. Umin = −e2E2/(2k) = −1 2αE2 . Physical check The factor one-half arises because the dipole is induced and requires internal deformation energy. The model is valid only for small displacements in a linear restoring regime.

Question 8

A displaced electron cloud gives a simple atomic polarizability

In a supplied classical response model, an electron cloud of charge −e-e is displaced a distance x opposite a uniform electric field of magnitude ℰ\mathcal E. A restoring force of magnitude kx acts toward the nucleus. Define the dipole magnitude p=exp=ex and p=αℰp=\alpha\mathcal E.

Diagram shown in the original SVG worksheet.

  1. Derive the equilibrium displacement and polarizability.

  2. Include the restoring energy and field coupling to find the minimum total energy.

Question 8 - Solution

A displaced electron cloud gives a simple atomic polarizability

Diagram shown in the original SVG worksheet.

(a) Method and calculation

kx=eℰ⇒x=eℰ/k.kx=e\mathcal E\Rightarrow\boxed{x=e\mathcal E/k}.p=ex=(e2/k)ℰ⇒α=e2/k.p=ex=(e^2/k)\mathcal E\Rightarrow\boxed{\alpha=e^2/k}.

(b) Interpretation and further calculation

U(x)=12kx2−eℰx,U′(x)=kx−eℰ=0.U(x)=\tfrac12kx^2-e\mathcal E x,\quad U\prime(x)=kx-e\mathcal E=0.Umin=−e2ℰ2/(2k)=−12αℰ2.\boxed{U_{\min}=-e^2\mathcal E^2/(2k)=-\tfrac12\alpha\mathcal E^2}.

Physical check

The factor one-half arises because the dipole is induced and requires internal deformation energy. The model is valid only for small displacements in a linear restoring regime.