Energy Bandgap of a Green LED: A green LED emits light at a wavelength of 5490 Å. Using Planck’s constant h = 6.6 × 10^-34 J·s, determine the approximate bandgap energy of the semiconductor material in electron volts (eV).

Difficulty: Medium

Correct Answer: 2.26 eV

Explanation:


Introduction / Context:
The emission wavelength of an LED corresponds directly to the bandgap energy (Eg) of its semiconductor material. Shorter wavelengths (blue, green) correspond to larger bandgaps compared to red or infrared LEDs.



Given Data / Assumptions:

  • λ = 5490 Å = 5.49 × 10^-7 m.
  • h = 6.6 × 10^-34 J·s.
  • c = 3 × 10^8 m/s.
  • 1 eV = 1.6 × 10^-19 J.


Concept / Approach:

Bandgap Eg = hc / λ. Convert from joules to eV.



Step-by-Step Solution:

Step 1: λ = 5.49 × 10^-7 m.Step 2: E = (6.6 × 10^-34 * 3 × 10^8) / (5.49 × 10^-7) ≈ 3.61 × 10^-19 J.Step 3: Eg = (3.61 × 10^-19) / (1.6 × 10^-19) ≈ 2.26 eV.


Verification / Alternative check:

Green emission ~2.2–2.4 eV, matches result.



Why Other Options Are Wrong:

  • 1.98 eV, 1.17 eV, 0.74 eV: longer wavelengths.
  • 3.10 eV: corresponds to violet.


Common Pitfalls:

  • Forgetting Å → m conversion.
  • Forgetting joule to eV conversion.


Final Answer:

2.26 eV

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