Question 5 of 6: Photon Emission in In-Doped Silicon and an Alumina Multilayer Capacitor

Question 5: Photon Emission in In-Doped Silicon and an Alumina Multilayer Capacitor (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

5.1 — Photon wavelengths from the In acceptor level in Si

Given. Silicon bandgap E_g = 1.07 eV; indium acceptor level (above the valence band) E_a = 0.16 eV; Planck's constant h = 4.136×10^-15 eV·s (the standard value; the source formula sheet's printed 4.375×10^-15 is not the physical constant, see the check note above); speed of light c = 3×10¹⁰ cm/s.

Find. The wavelength of the photon emitted when an electron (a) drops from the conduction band to the acceptor band, and (b) drops from the acceptor band to the valence band.

Approach. Each transition's photon energy is the corresponding energy gap; convert to wavelength with λ = hc/E.

1. (a) Conduction band → acceptor band. This transition spans the gap minus the acceptor level: $$E_{(a)}=E_g-E_a=1.07-0.16=0.91\ \text{eV}$$ $$\lambda_{(a)}=\dfrac{hc}{E_{(a)}}=\dfrac{(4.136\times10^{-15})(3\times10^{10})}{0.91}=1.364\times10^{-4}\ \text{cm}=\boxed{1364\ \text{nm}}\ \ (\text{near-infrared})$$
2. (b) Acceptor band → valence band. This transition spans only the acceptor level itself: $$\lambda_{(b)}=\dfrac{hc}{E_a}=\dfrac{(4.136\times10^{-15})(3\times10^{10})}{0.16}=7.755\times10^{-4}\ \text{cm}=\boxed{7755\ \text{nm}\ (7.76\ \mu\text{m})}\ \ (\text{mid-infrared})$$

If the formula sheet's printed h = 4.375×10^-15 eV·s is used instead, both wavelengths scale by 4.375/4.136 = 1.058, giving 1442 nm for (a) and 8203 nm for (b); the transition energies and their assignment are unchanged.

5.2 — Alumina multilayer capacitor sizing

Given. Sheet dimensions 1.5 cm × 1.5 cm × 0.001 cm; K = 6.5; ε₀ = 8.85×10^-14 F/cm; target capacitance C = 0.0142 µF.

Find. The number of sheets N needed.

Approach. A multilayer capacitor stacks sheets between alternating electrodes, so each sheet forms its own parallel-plate capacitor and the stack behaves as N identical capacitors connected in parallel: C_total = N·C₁.

1. Single-sheet capacitance. Sheet area A = 1.5 × 1.5 = 2.25 cm², thickness d = 0.001 cm: $$C_1=\dfrac{\varepsilon_0 K A}{d}=\dfrac{(8.85\times10^{-14})(6.5)(2.25)}{0.001}=1.294\times10^{-9}\ \text{F}=1.294\ \text{nF}$$
2. Sheets required. $$N=\dfrac{C}{C_1}=\dfrac{1.42\times10^{-8}}{1.294\times10^{-9}}=10.97\ \Rightarrow\ \boxed{N=11\ \text{sheets}}$$ (rounded up, since a fractional sheet cannot be built and 10 sheets would fall short of the target). This is the formula sheet's multilayer form C = Aε₀κ(n−1)/d read with n = number of electrode plates: n−1 = 10.97 → 11 alumina sheets interleaved between 12 plates.