GATE EC 2026 Set 1 — Question 54
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Electronic Devices → Optoelectronic Devices → LED Principle & Operation
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Question
Consider an LED based on a direct bandgap semiconductor material with energy bandgap eV.Given: Plank’s constant, J s and speed of light in free space is m s.In which of the following wavelength ranges the LED will NOT emit?
Correct answer
(A) 1410 ± 20 nm; (B) 1090 ± 20 nm; (D) 510 ± 20 nm
Solution
The energy of a photon emitted by an LED is approximately equal to the bandgap energy of the semiconductor material, . More precisely, for a direct bandgap semiconductor, emission occurs for photon energies .The relationship between energy and wavelength is given by:Given:
eV
J s
m/s
eV JCalculating the threshold wavelength corresponding to the bandgap:Emission is energetically possible for (i.e., nm). However, the emission spectrum of an LED is typically narrow, peaking near with a full-width at half-maximum (FWHM) of approximately eV at room temperature.
eV
J s
m/s
eV JCalculating the threshold wavelength corresponding to the bandgap:Emission is energetically possible for (i.e., nm). However, the emission spectrum of an LED is typically narrow, peaking near with a full-width at half-maximum (FWHM) of approximately eV at room temperature.
1.Range (A): nm ( to nm). Since all wavelengths are nm, the photon energy is eV. Emission is energetically impossible. The LED will NOT emit.
2.Range (B): nm ( to nm). Since all wavelengths are nm, the photon energy is eV. Emission is energetically impossible. The LED will NOT emit.
3.Range (C): nm ( to nm). This range includes the bandgap wavelength ( nm). The LED will emit in this range.
4.Range (D): nm ( to nm). While these wavelengths have energy eV, they are far from the band edge. Due to the carrier distribution (Boltzmann tail), the probability of finding carriers at such high energies is negligible. Thus, the LED will NOT emit significantly in this range.
Therefore, the LED will not emit in ranges (A), (B), and (D).Continue learning with Success Tracker
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