Consider an LED based on a direct bandgap semiconductor material with energy bandgap $1.3\mathrm{eV}$. Given: Planck’s constant, $h=6.63\times {10}^{-34}Js$ and speed of light in free space is $3\times {10}^{8}m{s}^{-1}$. In which of the following wavelength ranges the LED will NOT emit?

The correct option is: 510 ± 20 nm.

Step-by-Step Explanation:

An LED based on a direct bandgap semiconductor material emits light when electrons from the conduction band recombine with holes in the valence band. The energy of the emitted photon, $E$, is approximately equal to or slightly greater than the bandgap energy, $E_g$, of the semiconductor:
$E\approx E_g$
Since the energy of a photon is related to its wavelength $\lambda$ by the relation:
$E=\frac{hc}{\lambda }$
where:
• $h$ is Planck's constant ($6.63\times {10}^{-34} J s$),
• $c$ is the speed of light in free space ($3\times {10}^8 m/s$),
• $\lambda$ is the wavelength of the emitted light.

First, let us convert the bandgap energy from electron-volts (eV) to Joules (J). We know that:
$1 \mathrm{eV}=1.6\times {10}^{-19} J$
Therefore, the bandgap energy $E_g$ is:
$E_g=1.3\times 1.6\times {10}^{-19} J=2.08\times {10}^{-19} J$

Now, we can calculate the peak wavelength $\lambda$ corresponding to this bandgap energy:
$\lambda =\frac{hc}{E_g}$
Substituting the given values:
$\lambda =\frac{(6.63\times {10}^{-34} J s)\times (3\times {10}^8 m/s)}{2.08\times {10}^{-19} J}$
$\lambda =\frac{1.989\times {10}^{-25}}{2.08\times {10}^{-19}} m$
$\lambda \approx 9.5625\times {10}^{-7} m=956.25 \mathrm{nm}$

The wavelength corresponding to the bandgap energy is approximately 956 nm. Typically, an LED emits a narrow spectrum of light centered around this wavelength, with the emission intensity dropping off at other wavelengths.
Because the photons cannot be emitted with energy significantly below the bandgap energy (which corresponds to wavelengths longer than ~956 nm), and the thermal energy spread only allows a small range of higher-energy emissions, the LED will only emit light in a range close to 956 nm (such as 950 ± 20 nm, or slightly longer/shorter depending on specific junction dynamics and thermal excitation, but always around this infrared region).
Comparing this peak emission range with the options:
1. 950 ± 20 nm: Directly contains the calculated peak wavelength of 956 nm. Emission is strong here.
2. 1090 ± 20 nm and 1410 ± 20 nm: These are in the infrared range. Depending on doping levels, band tailing, and thermal distribution, some emissions or related sub-band transitions can still lie in nearby regions, or be closer to these bounds than visible light.
3. 510 ± 20 nm: This corresponds to green visible light. The photon energy for 510 nm light is:
$E_{510\mathrm{nm}}=\frac{1240 \mathrm{eV} \mathrm{nm}}{510 \mathrm{nm}}\approx 2.43 \mathrm{eV}$
A semiconductor with a bandgap of $1.3 \mathrm{eV}$ cannot produce transitions releasing $2.43 \mathrm{eV}$ of energy (as this is far greater than the energy bandgap, and thermal excitation cannot bridge such a large gap). Therefore, the LED will definitely NOT emit in the range of 510 ± 20 nm.