Semiconductors — Question 9

Section 41-5 · Physics pages 6888–6889

Question 9 An indirect optical transition must account for phonon energy An ideal indirect semiconductor has gap Eg = 1.10 eV. A phonon of energy Eph = .0400 eV supplies the required momentum. Ignore other losses and take hc = 1240 eV nm. k E γ + phonon Schematic; labels, not drawing lengths, specify the data. (a) Find photon thresholds for transitions accompanied by phonon absorption and phonon emission. (b) Find their threshold wavelengths and explain which process is suppressed as the phonon population tends to zero.
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Question 9 - Solution An indirect optical transition must account for phonon energy k E γ + phonon (a) Method and calculation Phonon absorption: Eγ + Eph = Eg ⇒Eγ,a = 1.06 eV . Phonon emission: Eγ = Eg + Eph ⇒Eγ,e = 1.14 eV . (b) Interpretation and further calculation λa = 1240/1.06 = 1170 nm, λe = 1240/1.14 = 1088 nm . Wa ∝Nph, We ∝Nph + 1 ⇒Wa →0 as Nph →0 . Physical check Emission can create a phonon even when none is initially present. The energy thresholds alone do not determine absorption strength or the complete spectrum.

Question 9

An indirect optical transition must account for phonon energy

An ideal indirect semiconductor has gap Eg=1.10eVE_g=1.10\,\mathrm{eV}. A phonon of energy Eph=.0400eVE_{\mathrm{ph}}=.0400\,\mathrm{eV} supplies the required momentum. Ignore other losses and take hc=1240eVnmhc=1240\,\mathrm{eV\,nm}.

Diagram shown in the original SVG worksheet.

  1. Find photon thresholds for transitions accompanied by phonon absorption and phonon emission.

  2. Find their threshold wavelengths and explain which process is suppressed as the phonon population tends to zero.

Question 9 - Solution

An indirect optical transition must account for phonon energy

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Phonon absorption: Eγ+Eph=Eg⇒Eγ,a=1.06eV.\text{Phonon absorption: }E_\gamma+E_{\mathrm{ph}}=E_g\Rightarrow\boxed{E_{\gamma,a}=1.06\,\mathrm{eV}}.Phonon emission: Eγ=Eg+Eph⇒Eγ,e=1.14eV.\text{Phonon emission: }E_\gamma=E_g+E_{\mathrm{ph}}\Rightarrow\boxed{E_{\gamma,e}=1.14\,\mathrm{eV}}.

(b) Interpretation and further calculation

λa=1240/1.06=1170nm,λe=1240/1.14=1088nm.\boxed{\lambda_a=1240/1.06=1170\,\mathrm{nm},\quad\lambda_e=1240/1.14=1088\,\mathrm{nm}}.Wa∝Nph,We∝Nph+1⇒Wa→0 as Nph→0.W_a\propto N_{\mathrm{ph}},\quad W_e\propto N_{\mathrm{ph}}+1\Rightarrow\boxed{W_a\to0\text{ as }N_{\mathrm{ph}}\to0}.

Physical check

Emission can create a phonon even when none is initially present. The energy thresholds alone do not determine absorption strength or the complete spectrum.