Consider an LED based on a direct bandgap semiconductor material with energy bandgap 1.3 eV. Given: ...

 

Q. Consider an LED based on a direct bandgap semiconductor material with energy bandgap 1.3 eV. Given: Plank’s constant, ℎ=6.63×10^−34 J s and speed of light in free space is 3×108 m s^−1. In which of the following wavelength ranges the LED will NOT emit?

Given Data

• Bandgap energy: Eg = 1.3 eV
• Planck's constant: h = 6.63 × 10⁻³⁴ J s
• Speed of light: c = 3 × 10⁸ m/s

Step 1: Convert Bandgap Energy into Joules

We know that:

1 eV = 1.602 × 10⁻¹⁹ J

Therefore,

Eg = 1.3 × 1.602 × 10⁻¹⁹

Eg = 2.0826 × 10⁻¹⁹ J

Step 2: Apply Planck's Formula

Photon energy is given by:

E = hc / λ

λ = hc / E

Substituting the values:

λ = (6.63 × 10⁻³⁴ × 3 × 10⁸)

/ (2.0826 × 10⁻¹⁹)

λ = 9.55 × 10⁻⁷ m

Step 3: Convert Metres into Nanometres

We know that 1 m = 10⁹ nm.

λ = 9.55 × 10⁻⁷ × 10⁹

λ ≈ 955 nm

Step 4: Understand the Concept

For an ideal direct bandgap LED, the emitted photon energy is approximately equal to the bandgap energy.

The maximum emission wavelength is approximately 955 nm. Wavelengths greater than this value have photon energy lower than the bandgap and cannot be emitted through band-to-band radiative recombination.

Step 5: Check the Options

Option	Wavelength Range	Result
A	1390–1430 nm	Not emitted
B	1070–1110 nm	Not emitted
C	930–970 nm	Allowed
D	490–530 nm	Not emitted

Answer: Options A, B, and D

Exam Shortcut Formula

λmax (nm) = 1240 / Eg (eV)

λmax = 1240 / 1.3

λmax ≈ 954 nm