Energies of a Trapped Electron — Question 1

Section 39-2 · Physics pages 6432–6433

Question 1 The first excitation probes an energy gap rather than the ground energy An electron occupies a well of width L = 1.00 nm. Model the electron by a one-dimensional infinite well: En = Cn2/L2, n = 1, 2, . . ., with C = .376 eV nm2 when L is in nanometers. Set the interior potential to zero. Use hc = 1240 eV nm and neglect recoil. n = 1 n = 2 n = 3 E Schematic; labels, not drawing lengths, specify the data. (a) Find the ground energy and the first excitation energy. (b) Find the photon wavelength resonant with the first excitation.
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Question 1 - Solution The first excitation probes an energy gap rather than the ground energy n = 1 n = 2 n = 3 E (a) Method and calculation E1 = C/L2 = .376 eV , E2 = 4E1. ∆E21 = 3E1 = 1.128 eV . (b) Interpretation and further calculation Eγ = hc/λ = ∆E21 ⇒λ = hc/(3E1). λ = 1099.29 nm . Physical check The photon supplies the difference between two levels. Using the ground-state energy itself as the absorption energy would identify the wrong wavelength.

Question 1

The first excitation probes an energy gap rather than the ground energy

An electron occupies a well of width L=1.00nmL=1.00\,\mathrm{nm}. Model the electron by a one-dimensional infinite well: En=Cn2/L2E_n=Cn^2/L^2, n=1,2,…n=1,2,\ldots, with C=.376eVnm2C=.376\,\mathrm{eV\,nm^2} when L is in nanometers. Set the interior potential to zero. Use hc=1240eVnmhc=1240\,\mathrm{eV\,nm} and neglect recoil.

Diagram shown in the original SVG worksheet.

  1. Find the ground energy and the first excitation energy.

  2. Find the photon wavelength resonant with the first excitation.

Question 1 - Solution

The first excitation probes an energy gap rather than the ground energy

Diagram shown in the original SVG worksheet.

(a) Method and calculation

E1=C/L2=.376eV,E2=4E1.\boxed{E_1=C/L^2=.376\,\mathrm{eV}},\quad E_2=4E_1.ΔE21=3E1=1.128eV.\boxed{\Delta E_{21}=3E_1=1.128\,\mathrm{eV}}.

(b) Interpretation and further calculation

Eγ=hc/λ=ΔE21⇒λ=hc/(3E1).E_\gamma=hc/\lambda=\Delta E_{21}\Rightarrow\lambda=hc/(3E_1).λ=1099.29nm.\boxed{\lambda=1099.29\,\mathrm{nm}}.

Physical check

The photon supplies the difference between two levels. Using the ground-state energy itself as the absorption energy would identify the wrong wavelength.