Official Correct Choice:
Option D: Blue photons have shorter wavelengths and higher energies, requiring wider-bandgap semiconductors (like GaN, \( E_g \approx 3.4\text{ eV} \)) with higher barrier potentials
Concept:The turn-on voltage \( V_F \) of an LED is approximately proportional to the emitted photon energy \( hf \) and the semiconductor bandgap: \( q V_F \approx E_g = \frac{h c}{\lambda} \).
Formula:$$V_F \approx \frac{E_g}{q} = \frac{h c}{q \lambda} \implies V_F \propto \frac{1}{\lambda}$$
Solution:- Red light has a longer wavelength (\( \lambda \approx 650\text{ nm} \)) and lower photon energy (\( \approx 1.9\text{ eV} \)), requiring \( V_F \approx 1.8\text{–}2.0\text{ V} \).
- Blue light has a shorter wavelength (\( \lambda \approx 450\text{ nm} \)) and higher photon energy (\( \approx 2.8\text{ eV} \)), requiring wider-bandgap materials (like GaN/InGaN, \( E_g \approx 3.4\text{ eV} \)) and higher forward voltages (\( V_F \approx 3.0\text{–}3.6\text{ V} \)).
Why other options are incorrect:- Option A: Blue LEDs are made of wide-bandgap semiconductors (GaN), not metals.
- Option B: LEDs emit light under forward bias, not reverse avalanche breakdown.
- Option C: Turn-on voltage is determined by the material's bandgap energy, not total electron count.
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