Question:medium
Consider an LED based on a direct bandgap semiconductor material with energy bandgap 1.3 eV.
Given: Planck's constant, \(h=6.63\times10^{-34}\) J s and speed of light in free space is \(3\times10^{8}\) m s\(^{-1}\).
In which of the following wavelength ranges the LED will NOT emit?
Consider an LED based on a direct bandgap semiconductor material with energy bandgap 1.3 eV.
Given: Planck's constant, \(h=6.63\times10^{-34}\) J s and speed of light in free space is \(3\times10^{8}\) m s\(^{-1}\).
In which of the following wavelength ranges the LED will NOT emit?
Given: Planck's constant, \(h=6.63\times10^{-34}\) J s and speed of light in free space is \(3\times10^{8}\) m s\(^{-1}\).
In which of the following wavelength ranges the LED will NOT emit?
Show Hint
An LED emits photons with energy close to its bandgap, not far above or below it.
Updated On: Jul 20, 2026
- \(1410\pm20\) nm
- \(1090\pm20\) nm
- \(950\pm20\) nm
- \(510\pm20\) nm
Show Solution
The Correct Option is A, B, D
Solution and Explanation
Step 1: Turn the question around, ask what energy each wavelength represents.
Rather than convert the bandgap to a wavelength first, check the photon energy of each option directly with $E=\dfrac{hc}{\lambda}$, using the combination $hc=6.63\times10^{-34}\times3\times10^8=1.989\times10^{-25}$ J·m, and remembering $1$ eV $=1.6\times10^{-19}$ J.
Step 2: Work out the energy for option (A), $1410$ nm.
\[ E=\frac{1.989\times10^{-25}}{1410\times10^{-9}}=1.41\times10^{-19}\ \text{J}=0.88\ \text{eV} \]
This is well below the $1.3$ eV bandgap, so a photon this small cannot come from band-to-band recombination.
Step 3: Work out the energy for option (B), $1090$ nm.
\[ E=\frac{1.989\times10^{-25}}{1090\times10^{-9}}=1.82\times10^{-19}\ \text{J}=1.14\ \text{eV} \]
Still below $1.3$ eV, so this too cannot be emitted.
Step 4: Work out the energy for option (C), $950$ nm.
\[ E=\frac{1.989\times10^{-25}}{950\times10^{-9}}=2.09\times10^{-19}\ \text{J}=1.31\ \text{eV} \]
This lands right on the bandgap of $1.3$ eV, exactly where recombination photons are expected.
Step 5: Work out the energy for option (D), $510$ nm.
\[ E=\frac{1.989\times10^{-25}}{510\times10^{-9}}=3.90\times10^{-19}\ \text{J}=2.44\ \text{eV} \]
This is almost double the bandgap. A single band-edge recombination event in this material has no way to give out that much energy.
Step 6: Conclude.
Options (A), (B) and (D) all sit at photon energies the material's bandgap cannot produce, so the LED will NOT emit in these three ranges, while (C) is where it does emit.
\[ \boxed{\text{(A), (B) and (D) are the wavelengths where the LED does NOT emit}} \]
Rather than convert the bandgap to a wavelength first, check the photon energy of each option directly with $E=\dfrac{hc}{\lambda}$, using the combination $hc=6.63\times10^{-34}\times3\times10^8=1.989\times10^{-25}$ J·m, and remembering $1$ eV $=1.6\times10^{-19}$ J.
Step 2: Work out the energy for option (A), $1410$ nm.
\[ E=\frac{1.989\times10^{-25}}{1410\times10^{-9}}=1.41\times10^{-19}\ \text{J}=0.88\ \text{eV} \]
This is well below the $1.3$ eV bandgap, so a photon this small cannot come from band-to-band recombination.
Step 3: Work out the energy for option (B), $1090$ nm.
\[ E=\frac{1.989\times10^{-25}}{1090\times10^{-9}}=1.82\times10^{-19}\ \text{J}=1.14\ \text{eV} \]
Still below $1.3$ eV, so this too cannot be emitted.
Step 4: Work out the energy for option (C), $950$ nm.
\[ E=\frac{1.989\times10^{-25}}{950\times10^{-9}}=2.09\times10^{-19}\ \text{J}=1.31\ \text{eV} \]
This lands right on the bandgap of $1.3$ eV, exactly where recombination photons are expected.
Step 5: Work out the energy for option (D), $510$ nm.
\[ E=\frac{1.989\times10^{-25}}{510\times10^{-9}}=3.90\times10^{-19}\ \text{J}=2.44\ \text{eV} \]
This is almost double the bandgap. A single band-edge recombination event in this material has no way to give out that much energy.
Step 6: Conclude.
Options (A), (B) and (D) all sit at photon energies the material's bandgap cannot produce, so the LED will NOT emit in these three ranges, while (C) is where it does emit.
\[ \boxed{\text{(A), (B) and (D) are the wavelengths where the LED does NOT emit}} \]
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