A light emitting diode (LED) is fabricated using GaAs semiconducting material whose band gap is 1.42 eV. The wavelength of light emitted from the LED is :

Solution

Physics sir · 4 steps

In an LED, one photon carries away exactly the band gap energy, so set hc/λ = E_g.

1. Photon from recombination. In an LED, an electron drops from the conduction band to the valence band and gives up energy equal to the band gap. That energy leaves as one photon.

   λ = hc/E_g

2. Put in numbers. hc = 6.63 × 10⁻³⁴ × 3 × 10⁸ = 1.989 × 10⁻²⁵ J m. The gap is E_g = 1.42 × 1.6 × 10⁻¹⁹ = 2.272 × 10⁻¹⁹ J.
3. Divide. λ = (1.989 × 10⁻²⁵)/(2.272 × 10⁻¹⁹) = 8.75 × 10⁻⁷ m ≈ 875 nm. The shortcut λ (nm) ≈ 1240/(E (eV)) gives about 873 nm, which is closest to the same option.
4. So. The GaAs LED emits near-infrared light of wavelength about 875 nm, option (b).

The idea to keep. Remember λ (nm) ≈ 1240 / E (eV). A bigger band gap gives a shorter wavelength. Use it for any LED, photodiode or photon-energy question.

Answer from the NTA final answer key. Solution by PrepWiser.