Light Emitting Diode
A light emitting diode is fabricated from a semiconductor whose energy band gap is 2.0 eV, and it emits photons whose energy equals this gap during recombination. What is the approximate wavelength of the emitted light?
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About This Question
- Subject
- physics
- Chapter
- semiconductor electronics
- Topic
- light emitting diode
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
620 nm
A light emitting diode emits light when electrons injected into the junction recombine with holes, releasing the energy difference as photons. In an ideal LED the photon energy equals the band gap, so (E_g = h\nu = hc/\lambda). Rearranging gives the wavelength (\lambda = hc/E_g). Using the convenient relation that (hc \approx 1240,\text{eV nm}), the wavelength is (\lambda = 1240 / 2.0 = 620,\text{nm}), which lies in the orange-red part of the visible spectrum. This inverse relationship between band gap and wavelength explains why different LED materials emit different colours: a wider gap pushes the emission toward blue and violet, while a narrower gap shifts it into the red and infrared. The value 310 nm comes from doubling the energy or halving the constant, placing it wrongly in the ultraviolet. The value 1240 nm results from omitting the band gap value entirely, giving an infrared answer. The value 414 nm corresponds to a 3.0 eV gap, not the stated 2.0 eV. As a final check, a 2.0 eV gap should yield visible light, and 620 nm sits squarely within the 400-700 nm visible range, confirming the calculation is physically reasonable.
This medium difficulty physics question is from the chapter semiconductor electronics, covering the topic of light emitting diode. It appeared in the 2025 exam.
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