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 \(T=295\ \mathrm K\), \(k_B T=0.026\ \mathrm{eV}\), and

\[h=6.62607015\times10^{-34}\ \mathrm{J\,s},\qquad c=2.99792458\times10^8\ \mathrm{m/s}.\]

Quick results

Problem	Result
2.1	Conduction-band separation: \(0.838\ \mathrm{eV}\)
2.2	Phonon: \(\lambda\approx1.25\ \mathrm{nm}\), \(f\approx6.83\ \mathrm{THz}\), \(E\approx28.2\ \mathrm{meV}\)
2.3	Electron wavelength: \(29.8\ \mathrm{nm}\), about 53 conventional cells and \(6.2\times10^5\) atoms
2.4	Correct relation: \(E(\mathrm{eV})=1239.84/\lambda(\mathrm{nm})\)
2.5	The 200–2000 nm interval corresponds to 6.20–0.620 eV
2.6	Free-particle equation: \(-\hbar^2\psi''/(2m)=E\psi\)
2.7	Ideally, reflection and transmission; no band-to-band absorption
2.8	Frequency is unchanged; wavelength and speed both fall by \(1/n\)

Problem 2.1: Energy step across a p-n junction

Paraphrase. At equilibrium, the electron densities on the two sides are \(n_n=10^{18}\ \mathrm{cm^{-3}}\) and \(n_p=10^4\ \mathrm{cm^{-3}}\). Find the conduction-band energy difference at room temperature.

For two electron populations in thermal equilibrium, the Boltzmann relation is

(1)\[\frac{n_p}{n_n} =\exp\left(-\frac{\Delta E_C}{k_B T}\right).\]

Take the natural logarithm and solve for \(\Delta E_C\):

\[\begin{split}\begin{aligned} \Delta E_C &=k_B T\ln\left(\frac{n_n}{n_p}\right)\\ &=(0.026\ \mathrm{eV}) \ln\left(\frac{10^{18}}{10^4}\right)\\ &=(0.026)(14\ln10)\ \mathrm{eV}\\ &=0.838\ \mathrm{eV}. \end{aligned}\end{split}\]

\[\boxed{\Delta E_C\approx0.84\ \mathrm{eV}}\]

The side with fewer conduction electrons has the higher conduction-band edge. As a check, inserting \(0.838\ \mathrm{eV}\) into Equation (1) returns the required density ratio \(10^{-14}\).

Problem 2.2: Photon-electron-phonon collision

Paraphrase. A \(1\ \mathrm{eV}\) 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 \(m_e=9.109\times10^{-31}\ \mathrm{kg}\), consistent with the chapter’s preceding \(1\ \mathrm{eV}\) electron estimate. The photon momentum,

\[p_\gamma=\frac{E_\gamma}{c}=5.34\times10^{-28}\ \mathrm{kg\,m/s},\]

is only about one thousandth of the final electron momentum.

At \(T=0\), no thermal phonon is available for absorption, so the physical branch is phonon emission. Neglecting the very small \(p_\gamma\) in the first estimate, momentum and energy conservation give

\[p_{\mathrm{ph}}\approx p_e=p,\]

(2)\[E_\gamma=\frac{p^2}{2m_e}+v_s p,\]

where \(v_s=8.5\times10^3\ \mathrm{m/s}\) and \(E_{\mathrm{ph}}=v_s p\).

Solve the quadratic

\[p^2+2m_e v_s p-2m_eE_\gamma=0\]

using the positive root:

\[p=m_e\left[ -v_s+\sqrt{v_s^2+\frac{2E_\gamma}{m_e}} \right] =5.33\times10^{-25}\ \mathrm{kg\,m/s}.\]

Phonon properties

The phonon de Broglie wavelength is

\[\lambda_{\mathrm{ph}} =\frac{h}{p_{\mathrm{ph}}} \approx\frac{6.626\times10^{-34}}{5.32\times10^{-25}} =1.25\times10^{-9}\ \mathrm m.\]

Its frequency and energy are

\[f_{\mathrm{ph}} =\frac{v_s}{\lambda_{\mathrm{ph}}} =\frac{8.5\times10^3}{1.25\times10^{-9}} \approx6.83\times10^{12}\ \mathrm{Hz},\]

\[E_{\mathrm{ph}}=hf_{\mathrm{ph}} \approx4.52\times10^{-21}\ \mathrm J =0.0282\ \mathrm{eV}.\]

Thus,

\[\boxed{ \lambda_{\mathrm{ph}}\approx1.25\ \mathrm{nm},\quad f_{\mathrm{ph}}\approx6.83\ \mathrm{THz},\quad E_{\mathrm{ph}}\approx28.2\ \mathrm{meV} }.\]

Final and room-temperature electron energies

For the phonon-emission branch, Equation (2) gives

\[E_{e,f}=E_\gamma-E_{\mathrm{ph}} =1.000-0.0282 =\boxed{0.972\ \mathrm{eV}}.\]

If a phonon is already present and is absorbed instead, the corresponding solution is approximately \(E_{e,f}=1.029\ \mathrm{eV}\). Stating the phonon branch is therefore essential.

At room temperature the characteristic initial thermal energy is

\[E_{\mathrm{thermal}}\sim k_BT\approx0.026\ \mathrm{eV}.\]

The three-dimensional mean translational energy is \(3k_BT/2\approx0.039\ \mathrm{eV}\). 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 \(m^*=0.065m_e\) and thermal kinetic energy \(k_BT\) 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,

\[E=\frac{p^2}{2m^*},\qquad \lambda=\frac{h}{p},\]

so

\[\lambda =\frac{h}{\sqrt{2m^*E}} =\frac{6.626\times10^{-34}} {\sqrt{2(0.065)(9.109\times10^{-31}) (0.026)(1.602\times10^{-19})}}.\]

Therefore,

\[\boxed{\lambda\approx2.98\times10^{-8}\ \mathrm m=29.8\ \mathrm{nm}}.\]

Crystal cells along the wavelength

The problem does not provide a lattice constant. Using the standard room-temperature GaAs conventional-cell dimension \(a=0.565\ \mathrm{nm}\),

\[N_{\mathrm{cells}}=\frac{\lambda}{a} =\frac{29.8}{0.565}=52.8.\]

The wavelength spans approximately

\[\boxed{53\ \text{conventional unit cells}}.\]

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 \(8/a^3\). A sphere of diameter \(\lambda\) has volume \(\pi\lambda^3/6\), hence

\[\begin{split}\begin{aligned} N_{\mathrm{atoms}} &=\frac{\pi\lambda^3}{6}\frac{8}{a^3}\\ &=\frac{4\pi}{3}\left(\frac{\lambda}{a}\right)^3\\ &=\frac{4\pi}{3}(52.8)^3\\ &\approx6.16\times10^5. \end{aligned}\end{split}\]

\[\boxed{N_{\mathrm{atoms}}\approx6.2\times10^5\ \text{atoms}}.\]

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 \(E=hf\) and \(c=f\lambda\).

Eliminate frequency:

\[E=\frac{hc}{\lambda}.\]

For wavelength in nanometres, write \(\lambda=\lambda_{\mathrm{nm}}10^{-9}\ \mathrm m\), then convert joules to electron volts:

\[\begin{split}\begin{aligned} E(\mathrm{eV}) &=\frac{(6.62607015\times10^{-34}\ \mathrm{J\,s}) (2.99792458\times10^8\ \mathrm{m/s})} {(\lambda_{\mathrm{nm}}10^{-9}\ \mathrm m) (1.602176634\times10^{-19}\ \mathrm{J/eV})}\\ &=\frac{1239.841984}{\lambda_{\mathrm{nm}}}\ \mathrm{eV}. \end{aligned}\end{split}\]

Thus the convenient rounded relation is

(3)\[\boxed{ E(\mathrm{eV}) \approx\frac{1240}{\lambda(\mathrm{nm})} }.\]

Important

Typographical error in the problem

The formula printed in Problem 2.4 has \(124\) in the numerator. It is missing a zero. The chapter’s own earlier result that a \(1\ \mathrm{eV}\) photon has wavelength \(1240\ \mathrm{nm}\) confirms the correct constant.

Problem 2.5: Energy-wavelength conversion chart

Paraphrase. Construct aligned wavelength and photon-energy axes from \(200\) to \(2000\ \mathrm{nm}\); mark blue, green, red, and the \(1550\ \mathrm{nm}\) telecommunications region.

Use Equation (3) at the two endpoints:

\[E(200\ \mathrm{nm})=\frac{1240}{200}=6.20\ \mathrm{eV},\]

\[E(2000\ \mathrm{nm})=\frac{1240}{2000}=0.620\ \mathrm{eV}.\]

The requested corresponding energy interval is therefore

\[\boxed{0.620\ \mathrm{eV}\le E\le6.20\ \mathrm{eV}}.\]

A wavelength-linear conversion chart. The upper energy labels are nonlinear because \(E\) is proportional to \(1/\lambda\). 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,

\[E(1550\ \mathrm{nm})=\frac{1240}{1550}=0.800\ \mathrm{eV}.\]

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

\[\psi(x)=A\sin(kx).\]

The two derivatives are

\[\frac{d\psi}{dx}=Ak\cos(kx),\]

\[\frac{d^2\psi}{dx^2} =-Ak^2\sin(kx) =\boxed{-k^2\psi(x)}.\]

Part b: introduce momentum and energy. Since \(k=2\pi/\lambda\) and \(\hbar=h/(2\pi)\), de Broglie’s relation gives

\[p=\frac{h}{\lambda}=\hbar k.\]

The nonrelativistic kinetic energy is therefore

\[E=\frac{p^2}{2m}=\frac{\hbar^2k^2}{2m}.\]

Multiply the second-derivative equation by \(-\hbar^2/(2m)\):

\[-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} =\frac{\hbar^2k^2}{2m}\psi =E\psi.\]

Thus,

(4)\[\boxed{ -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}=E\psi }.\]

Equation (4) is the one-dimensional, time-independent Schrödinger equation for a free particle. A potential \(V(x)\) adds a term \(V(x)\psi(x)\) on the left.

Problem 2.7: Sub-bandgap light incident on silicon

Paraphrase. Decide whether \(1240\ \mathrm{nm}\) light is absorbed, reflected, or transmitted by a \(0.5\ \mathrm{mm}\) silicon wafer whose band gap is \(1.1\ \mathrm{eV}\).

The photon energy is

\[E_\gamma=\frac{1240}{1240}\ \mathrm{eV}=1.00\ \mathrm{eV}.\]

Since

\[E_\gamma=1.00\ \mathrm{eV}<E_g=1.1\ \mathrm{eV},\]

one photon cannot promote a valence electron across the band gap. In the ideal model there is therefore no band-to-band absorption. The air-silicon index discontinuity still reflects part of the beam, and the remainder is transmitted through the wafer:

\[\boxed{\text{reflection and transmission occur; intrinsic absorption does not.}}\]

Real wafers can have weak free-carrier, defect, surface, or multiphoton absorption. Those mechanisms are outside the three-process idealization in the problem.

Problem 2.8: What changes when light enters glass?

Paraphrase. Light crosses from air into glass of refractive index \(n=1.5\). Determine whether its frequency, wavelength, or both change, and justify the result using photon energy.

A stationary boundary cannot create a new time oscillation rate. Equivalently, the photon energy is conserved across a passive interface:

\[E_1=E_2,\qquad hf_1=hf_2,\]

so

\[\boxed{f_2=f_1}.\]

The speed in glass is

\[v_2=\frac{c}{n}=\frac{c}{1.5}.\]

Because \(v=f\lambda\) and the frequency is unchanged,

\[\lambda_2=\frac{v_2}{f} =\frac{c}{nf} =\frac{\lambda_1}{n} =\frac{2}{3}\lambda_1.\]

Therefore the speed and wavelength both decrease by the factor \(1/n\), while frequency and photon energy remain unchanged:

\[\boxed{ v_{\mathrm{glass}}=\frac{2}{3}c,\qquad \lambda_{\mathrm{glass}}=\frac{2}{3}\lambda_{\mathrm{air}},\qquad f_{\mathrm{glass}}=f_{\mathrm{air}} }.\]