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Electronic Devices & CircuitsOptoelectronic Devices

The emitted photon energy from an LED is approximately related to wavelength lambda by?

  • AE=hc/lambda
  • BE=h lambda/c
  • CE=c/(h lambda)
  • DE=q lambda only
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Correct Answer: A. E=hc/lambda
Explanation: Photon energy is E=hf=hc/lambda, so shorter wavelengths correspond to higher photon energies.
Reference: Standard optoelectronics: photon energy
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