Photonics Essentials: Chapter 2 Problems ======================================== Source ------ Thomas P. Pearsall, *Photonics Essentials: An Introduction with Experiments* (McGraw-Hill, 2003), Chapter 2, ``Electrons and Photons``, Problems 2.1--2.8, printed pages 32--34. The problems are paraphrased below. Calculations use :math:`T=295\ \mathrm K`, :math:`k_B T=0.026\ \mathrm{eV}`, and .. math:: h=6.62607015\times10^{-34}\ \mathrm{J\,s},\qquad c=2.99792458\times10^8\ \mathrm{m/s}. Quick results ------------- .. csv-table:: :header: "Problem", "Result" "2.1", "Conduction-band separation: :math:`0.838\ \mathrm{eV}`" "2.2", "Phonon: :math:`\lambda\approx1.25\ \mathrm{nm}`, :math:`f\approx6.83\ \mathrm{THz}`, :math:`E\approx28.2\ \mathrm{meV}`" "2.3", "Electron wavelength: :math:`29.8\ \mathrm{nm}`, about 53 conventional cells and :math:`6.2\times10^5` atoms" "2.4", "Correct relation: :math:`E(\mathrm{eV})=1239.84/\lambda(\mathrm{nm})`" "2.5", "The 200--2000 nm interval corresponds to 6.20--0.620 eV" "2.6", "Free-particle equation: :math:`-\hbar^2\psi''/(2m)=E\psi`" "2.7", "Ideally, reflection and transmission; no band-to-band absorption" "2.8", "Frequency is unchanged; wavelength and speed both fall by :math:`1/n`" Problem 2.1: Energy step across a p-n junction ------------------------------------------------ **Paraphrase.** At equilibrium, the electron densities on the two sides are :math:`n_n=10^{18}\ \mathrm{cm^{-3}}` and :math:`n_p=10^4\ \mathrm{cm^{-3}}`. Find the conduction-band energy difference at room temperature. For two electron populations in thermal equilibrium, the Boltzmann relation is .. math:: :label: pearsall-boltzmann-ratio \frac{n_p}{n_n} =\exp\left(-\frac{\Delta E_C}{k_B T}\right). Take the natural logarithm and solve for :math:`\Delta E_C`: .. math:: \begin{aligned} \Delta E_C &=k_B T\ln\left(\frac{n_n}{n_p}\right)\\ &=(0.026\ \mathrm{eV}) \ln\left(\frac{10^{18}}{10^4}\right)\\ &=(0.026)(14\ln10)\ \mathrm{eV}\\ &=0.838\ \mathrm{eV}. \end{aligned} .. math:: \boxed{\Delta E_C\approx0.84\ \mathrm{eV}} The side with fewer conduction electrons has the higher conduction-band edge. As a check, inserting :math:`0.838\ \mathrm{eV}` into Equation :eq:`pearsall-boltzmann-ratio` returns the required density ratio :math:`10^{-14}`. Problem 2.2: Photon-electron-phonon collision --------------------------------------------- **Paraphrase.** A :math:`1\ \mathrm{eV}` photon transfers energy to an electron initially at rest. A silicon phonon supplies the momentum balance. Find the phonon wavelength, frequency, and energy; then find the electron energy and discuss the room-temperature initial state. Assumptions and conservation laws ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ The problem does not specify an electron effective mass, so we use the free electron mass :math:`m_e=9.109\times10^{-31}\ \mathrm{kg}`, consistent with the chapter's preceding :math:`1\ \mathrm{eV}` electron estimate. The photon momentum, .. math:: p_\gamma=\frac{E_\gamma}{c}=5.34\times10^{-28}\ \mathrm{kg\,m/s}, is only about one thousandth of the final electron momentum. At :math:`T=0`, no thermal phonon is available for absorption, so the physical branch is **phonon emission**. Neglecting the very small :math:`p_\gamma` in the first estimate, momentum and energy conservation give .. math:: p_{\mathrm{ph}}\approx p_e=p, .. math:: :label: pearsall-photon-phonon-energy E_\gamma=\frac{p^2}{2m_e}+v_s p, where :math:`v_s=8.5\times10^3\ \mathrm{m/s}` and :math:`E_{\mathrm{ph}}=v_s p`. Solve the quadratic .. math:: p^2+2m_e v_s p-2m_eE_\gamma=0 using the positive root: .. math:: p=m_e\left[ -v_s+\sqrt{v_s^2+\frac{2E_\gamma}{m_e}} \right] =5.33\times10^{-25}\ \mathrm{kg\,m/s}. Phonon properties ~~~~~~~~~~~~~~~~~ The phonon de Broglie wavelength is .. math:: \lambda_{\mathrm{ph}} =\frac{h}{p_{\mathrm{ph}}} \approx\frac{6.626\times10^{-34}}{5.32\times10^{-25}} =1.25\times10^{-9}\ \mathrm m. Its frequency and energy are .. math:: f_{\mathrm{ph}} =\frac{v_s}{\lambda_{\mathrm{ph}}} =\frac{8.5\times10^3}{1.25\times10^{-9}} \approx6.83\times10^{12}\ \mathrm{Hz}, .. math:: E_{\mathrm{ph}}=hf_{\mathrm{ph}} \approx4.52\times10^{-21}\ \mathrm J =0.0282\ \mathrm{eV}. Thus, .. math:: \boxed{ \lambda_{\mathrm{ph}}\approx1.25\ \mathrm{nm},\quad f_{\mathrm{ph}}\approx6.83\ \mathrm{THz},\quad E_{\mathrm{ph}}\approx28.2\ \mathrm{meV} }. Final and room-temperature electron energies ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ For the phonon-emission branch, Equation :eq:`pearsall-photon-phonon-energy` gives .. math:: E_{e,f}=E_\gamma-E_{\mathrm{ph}} =1.000-0.0282 =\boxed{0.972\ \mathrm{eV}}. If a phonon is already present and is **absorbed** instead, the corresponding solution is approximately :math:`E_{e,f}=1.029\ \mathrm{eV}`. Stating the phonon branch is therefore essential. At room temperature the characteristic initial thermal energy is .. math:: E_{\mathrm{thermal}}\sim k_BT\approx0.026\ \mathrm{eV}. The three-dimensional mean translational energy is :math:`3k_BT/2\approx0.039\ \mathrm{eV}`. The exact initial energy and momentum are thermally distributed, so a room-temperature collision does not have one unique initial value. Problem 2.3: Thermal electron wavelength in GaAs ------------------------------------------------- **Paraphrase.** Use the GaAs electron effective mass :math:`m^*=0.065m_e` and thermal kinetic energy :math:`k_BT` to find its de Broglie wavelength, express that length in crystal cells, and estimate how many atoms occupy a sphere of that diameter. Electron wavelength ~~~~~~~~~~~~~~~~~~~ For a nonrelativistic electron, .. math:: E=\frac{p^2}{2m^*},\qquad \lambda=\frac{h}{p}, so .. math:: \lambda =\frac{h}{\sqrt{2m^*E}} =\frac{6.626\times10^{-34}} {\sqrt{2(0.065)(9.109\times10^{-31}) (0.026)(1.602\times10^{-19})}}. Therefore, .. math:: \boxed{\lambda\approx2.98\times10^{-8}\ \mathrm m=29.8\ \mathrm{nm}}. Crystal cells along the wavelength ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ The problem does not provide a lattice constant. Using the standard room-temperature GaAs conventional-cell dimension :math:`a=0.565\ \mathrm{nm}`, .. math:: N_{\mathrm{cells}}=\frac{\lambda}{a} =\frac{29.8}{0.565}=52.8. The wavelength spans approximately .. math:: \boxed{53\ \text{conventional unit cells}}. Atoms in a wavelength-diameter sphere ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ The conventional zinc-blende GaAs cell contains four Ga atoms and four As atoms, or eight atoms total. The atomic number density in this cell model is :math:`8/a^3`. A sphere of diameter :math:`\lambda` has volume :math:`\pi\lambda^3/6`, hence .. math:: \begin{aligned} N_{\mathrm{atoms}} &=\frac{\pi\lambda^3}{6}\frac{8}{a^3}\\ &=\frac{4\pi}{3}\left(\frac{\lambda}{a}\right)^3\\ &=\frac{4\pi}{3}(52.8)^3\\ &\approx6.16\times10^5. \end{aligned} .. math:: \boxed{N_{\mathrm{atoms}}\approx6.2\times10^5\ \text{atoms}}. This large number illustrates what it means for a conduction electron to be delocalized over the crystal. Problem 2.4: Photon energy from wavelength ------------------------------------------- **Paraphrase.** Derive the electron-volt photon-energy formula from :math:`E=hf` and :math:`c=f\lambda`. Eliminate frequency: .. math:: E=\frac{hc}{\lambda}. For wavelength in nanometres, write :math:`\lambda=\lambda_{\mathrm{nm}}10^{-9}\ \mathrm m`, then convert joules to electron volts: .. math:: \begin{aligned} E(\mathrm{eV}) &=\frac{(6.62607015\times10^{-34}\ \mathrm{J\,s}) (2.99792458\times10^8\ \mathrm{m/s})} {(\lambda_{\mathrm{nm}}10^{-9}\ \mathrm m) (1.602176634\times10^{-19}\ \mathrm{J/eV})}\\ &=\frac{1239.841984}{\lambda_{\mathrm{nm}}}\ \mathrm{eV}. \end{aligned} Thus the convenient rounded relation is .. math:: :label: pearsall-photon-energy \boxed{ E(\mathrm{eV}) \approx\frac{1240}{\lambda(\mathrm{nm})} }. .. important:: Typographical error in the problem The formula printed in Problem 2.4 has :math:`124` in the numerator. It is missing a zero. The chapter's own earlier result that a :math:`1\ \mathrm{eV}` photon has wavelength :math:`1240\ \mathrm{nm}` confirms the correct constant. Problem 2.5: Energy-wavelength conversion chart ------------------------------------------------ **Paraphrase.** Construct aligned wavelength and photon-energy axes from :math:`200` to :math:`2000\ \mathrm{nm}`; mark blue, green, red, and the :math:`1550\ \mathrm{nm}` telecommunications region. Use Equation :eq:`pearsall-photon-energy` at the two endpoints: .. math:: E(200\ \mathrm{nm})=\frac{1240}{200}=6.20\ \mathrm{eV}, .. math:: E(2000\ \mathrm{nm})=\frac{1240}{2000}=0.620\ \mathrm{eV}. The requested corresponding energy interval is therefore .. math:: \boxed{0.620\ \mathrm{eV}\le E\le6.20\ \mathrm{eV}}. .. figure:: ../../../_static/knowledge_base/worked_exercises/photonics_essentials/ch02_energy_wavelength.svg :alt: Aligned wavelength and photon-energy scales from 200 to 2000 nanometres with blue, green, red, and 1550 nanometre regions marked :width: 100% :align: center A wavelength-linear conversion chart. The upper energy labels are nonlinear because :math:`E` is proportional to :math:`1/\lambda`. Colour boundaries are approximate and vary slightly among references. The chart uses approximate colour intervals of 450--495 nm for blue, 495--570 nm for green, and 620--700 nm for red. At the fibre telecommunications wavelength, .. math:: E(1550\ \mathrm{nm})=\frac{1240}{1550}=0.800\ \mathrm{eV}. Blue photons have more energy than red photons because blue has the shorter wavelength. Problem 2.6: From a sinusoidal wave to electron energy ------------------------------------------------------- **Part a: differentiate the wave.** Start with .. math:: \psi(x)=A\sin(kx). The two derivatives are .. math:: \frac{d\psi}{dx}=Ak\cos(kx), .. math:: \frac{d^2\psi}{dx^2} =-Ak^2\sin(kx) =\boxed{-k^2\psi(x)}. **Part b: introduce momentum and energy.** Since :math:`k=2\pi/\lambda` and :math:`\hbar=h/(2\pi)`, de Broglie's relation gives .. math:: p=\frac{h}{\lambda}=\hbar k. The nonrelativistic kinetic energy is therefore .. math:: E=\frac{p^2}{2m}=\frac{\hbar^2k^2}{2m}. Multiply the second-derivative equation by :math:`-\hbar^2/(2m)`: .. math:: -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} =\frac{\hbar^2k^2}{2m}\psi =E\psi. Thus, .. math:: :label: pearsall-free-schrodinger \boxed{ -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}=E\psi }. Equation :eq:`pearsall-free-schrodinger` is the one-dimensional, time-independent Schrödinger equation for a free particle. A potential :math:`V(x)` adds a term :math:`V(x)\psi(x)` on the left. Problem 2.7: Sub-bandgap light incident on silicon -------------------------------------------------- **Paraphrase.** Decide whether :math:`1240\ \mathrm{nm}` light is absorbed, reflected, or transmitted by a :math:`0.5\ \mathrm{mm}` silicon wafer whose band gap is :math:`1.1\ \mathrm{eV}`. The photon energy is .. math:: E_\gamma=\frac{1240}{1240}\ \mathrm{eV}=1.00\ \mathrm{eV}. Since .. math:: E_\gamma=1.00\ \mathrm{eV}