Electronic Devices · Study notes

PHYSemiconductors & Energy Bands

Energy Bands & Band Gap

The band gap $E_g$ is the energy separation between the top of the valence band and the bottom of the conduction band; an electron must absorb at least $E_g$ to cross it, and an LED emits a photon of energy $\approx E_g$ when an electron falls back. ISC numericals link $E_g$ to the longest absorbable wavelength or to the emitted LED wavelength, so the $E = hc/\lambda$ relation and the eV-to-joule conversion are the heart of this subtopic.

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Photon energy from wavelength

E = \dfrac{hc}{\lambda}

E

= photon energy,

h = 6.63 \times 10^{-34}\,\text{J\,s}

c = 3 \times 10^{8}\,\text{m\,s}^{-1}

\lambda

= wavelength in metres.

Maximum absorbable wavelength

\lambda_{max} = \dfrac{hc}{E_g}

E_g

= band gap (in joules). Longest wavelength that can still excite an electron across the gap.

Band gap from LED wavelength

E_g = \dfrac{hc}{\lambda}

\lambda

= emitted LED wavelength; the emitted photon energy approximately equals the band gap.

Energy unit conversion

E_{(\text{eV})} = \dfrac{E_{(\text{J})}}{1.6 \times 10^{-19}}

Converts joules to electron-volts;

1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}

Only photons with energy $\geq E_g$ can excite an electron across the gap, so $\lambda_{max} = hc/E_g$ is a cut-off (longest) wavelength.
Larger $E_g$ means shorter $\lambda_{max}$ — wide-gap materials respond to higher-frequency light.
An LED's emitted photon energy is close to $E_g$ , giving its characteristic colour.
Use $hc \approx 1240\,\text{eV\,nm}$ as a fast shortcut: $E_g(\text{eV}) = 1240/\lambda(\text{nm})$ .
Always convert $\text{nm}$ to metres ( $1\,\text{nm} = 10^{-9}\,\text{m}$ ) when using SI $h$ and $c$ .
Band gaps: Si $\approx 1.1\,\text{eV}$ , Ge $\approx 0.7\,\text{eV}$ ; insulators have $E_g > 3\,\text{eV}$ .
Answers in joules can be converted to eV by dividing by $1.6 \times 10^{-19}$ for comparison with standard band-gap values.

Where the marks go

Forgetting to convert wavelength from nm to m, giving an answer off by $10^{9}$ .
Reporting $E_g$ in joules when the question expects eV (or vice versa) — always divide/multiply by $1.6 \times 10^{-19}$ .
Confusing the cut-off as a minimum wavelength; $\lambda_{max}$ is the longest wavelength that still works.
Using $\lambda$ in $E = hf$ or frequency $f$ in $E = hc/\lambda$ — pick the form that matches the given quantity.

How the board asks it

Numerical $E=hc/\lambda$ for an LED
A light-emitting diode emits monochromatic light of wavelength $660\,\text{nm}$ . Calculate the energy band gap $E_g$ (in $\text{eV}$ ) of the semiconductor used to fabricate it. (Take $h = 6.6 \times 10^{-34}\,\text{J s}$ , $c = 3 \times 10^{8}\,\text{m s}^{-1}$ and $1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}$ .)
Numericalcut-off wavelength $\lambda_{max}=hc/E_g$
The energy band gap of a semiconductor is $1.24\,\text{eV}$ . Find the maximum wavelength of radiation that can be absorbed to excite an electron from the valence band to the conduction band.
Define / statevalence-to-conduction band separation
Define the term 'energy band gap' of a solid and state its approximate value for silicon.
Distinguishband gaps of conductor, semiconductor, insulator
Distinguish between a conductor, a semiconductor and an insulator on the basis of their energy band gaps, quoting a typical value of $E_g$ for each.
Give reasonslarger $E_g$ gives shorter $\lambda_{max}$
Explain why a wide-band-gap semiconductor can absorb only shorter-wavelength light than a narrow-band-gap one.

Practise this topic

NextIntrinsic & Extrinsic Semiconductors

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.