Question:
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
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The Correct Option is A, B, D
Solution and Explanation
Step 1: Recall how an LED emits light.
In a direct bandgap semiconductor, an electron in the conduction band recombines with a hole in the valence band and releases the energy difference as a photon. The photon energy released in this band-to-band recombination is very close to the bandgap energy \(E_g\), with only a small spread of a few \(kT\) around it coming from the thermal spread of carrier energies. An LED does not efficiently emit photons whose energy is far below \(E_g\) (no such transition is available) or far above \(E_g\) (very few carriers sit that high above the band edge).
Step 2: Convert the bandgap energy into a wavelength.
The photon energy and wavelength are related by
\[ E_g=\frac{hc}{\lambda_g}\ \Rightarrow\ \lambda_g=\frac{hc}{E_g} \]
Using \(h=6.63\times10^{-34}\ J\,s\), \(c=3\times10^8\ m\,s^{-1}\), and converting \(E_g=1.3\ eV\) to joules with \(1\ eV=1.6\times10^{-19}\ J\):
\[ E_g=1.3\times1.6\times10^{-19}=2.08\times10^{-19}\ J \]
\[ \lambda_g=\frac{6.63\times10^{-34}\times3\times10^8}{2.08\times10^{-19}}=\frac{1.989\times10^{-25}}{2.08\times10^{-19}}\approx9.56\times10^{-7}\ m=956\ nm \]
Step 3: Compare each option's range to \(\lambda_g\approx956\ nm\).
(A) \(1410\pm20\) nm: This is far longer than \(956\) nm, corresponding to a photon energy well below \(E_g\). Direct band-to-band recombination cannot supply photons with energy less than the bandgap, so the LED will NOT emit here.
(B) \(1090\pm20\) nm: Also longer than \(956\) nm (energy below \(E_g\)), so this too is below the smallest photon energy the recombination can give. The LED will NOT emit here either.
(C) \(950\pm20\) nm: This range straddles the calculated \(\lambda_g\approx956\) nm almost exactly, which is where band-edge recombination peaks. The LED DOES emit in this range.
(D) \(510\pm20\) nm: This corresponds to a photon energy of roughly \(2.4\) eV, nearly double \(E_g=1.3\) eV. Ordinary band-edge recombination in this material has no mechanism to release almost twice the bandgap energy as a single photon, so the LED will NOT emit here.
Step 4: Final conclusion.
The wavelength ranges where the LED will NOT emit are
\[ \boxed{1410\pm20\ nm,\ \ 1090\pm20\ nm,\ \ 510\pm20\ nm} \]
that is, options (A), (B) and (D).
In a direct bandgap semiconductor, an electron in the conduction band recombines with a hole in the valence band and releases the energy difference as a photon. The photon energy released in this band-to-band recombination is very close to the bandgap energy \(E_g\), with only a small spread of a few \(kT\) around it coming from the thermal spread of carrier energies. An LED does not efficiently emit photons whose energy is far below \(E_g\) (no such transition is available) or far above \(E_g\) (very few carriers sit that high above the band edge).
Step 2: Convert the bandgap energy into a wavelength.
The photon energy and wavelength are related by
\[ E_g=\frac{hc}{\lambda_g}\ \Rightarrow\ \lambda_g=\frac{hc}{E_g} \]
Using \(h=6.63\times10^{-34}\ J\,s\), \(c=3\times10^8\ m\,s^{-1}\), and converting \(E_g=1.3\ eV\) to joules with \(1\ eV=1.6\times10^{-19}\ J\):
\[ E_g=1.3\times1.6\times10^{-19}=2.08\times10^{-19}\ J \]
\[ \lambda_g=\frac{6.63\times10^{-34}\times3\times10^8}{2.08\times10^{-19}}=\frac{1.989\times10^{-25}}{2.08\times10^{-19}}\approx9.56\times10^{-7}\ m=956\ nm \]
Step 3: Compare each option's range to \(\lambda_g\approx956\ nm\).
(A) \(1410\pm20\) nm: This is far longer than \(956\) nm, corresponding to a photon energy well below \(E_g\). Direct band-to-band recombination cannot supply photons with energy less than the bandgap, so the LED will NOT emit here.
(B) \(1090\pm20\) nm: Also longer than \(956\) nm (energy below \(E_g\)), so this too is below the smallest photon energy the recombination can give. The LED will NOT emit here either.
(C) \(950\pm20\) nm: This range straddles the calculated \(\lambda_g\approx956\) nm almost exactly, which is where band-edge recombination peaks. The LED DOES emit in this range.
(D) \(510\pm20\) nm: This corresponds to a photon energy of roughly \(2.4\) eV, nearly double \(E_g=1.3\) eV. Ordinary band-edge recombination in this material has no mechanism to release almost twice the bandgap energy as a single photon, so the LED will NOT emit here.
Step 4: Final conclusion.
The wavelength ranges where the LED will NOT emit are
\[ \boxed{1410\pm20\ nm,\ \ 1090\pm20\ nm,\ \ 510\pm20\ nm} \]
that is, options (A), (B) and (D).
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