Question

For LED’s to emit light in the visible region of electromagnetic light, it should have an energy band gap in the range of:
A). \[\text{0}\text{.1eV to 0}\text{.4eV}\]
B). \[\text{0}\text{.5eV to 0}\text{.8eV}\]
C). \[\text{0}\text{.9eV to 1}\text{.6eV}\]
D). \[\text{1}\text{.7eV to 3}\text{.0eV}\]

Answer 1

Hint: The visible part of the electromagnetic spectrum belongs to the range of \[400\text{nm to 700nm}\]. The energy band gap is directly proportional to the frequency of light it emits. In order to find the energy band gap range, we need to calculate the energy bandgap for the lowest wavelength and the highest wavelength in the visible part of the electromagnetic spectrum.

Complete step by step answer:
So, the energy band gap of an LED in order to emit light in the visible spectrum depends on the frequency of the photon it emits when the electrons drop from the conduction to the valence band.
So, the bandgap of an LED can be expressed as a function of the wavelength of the photon emitted given by,
${{E}_{b}}=\dfrac{hc}{\lambda }$
Where
${{E}_{b}}$ is a is the energy bandgap of LED
h is the Planck’s Constant whose value is $6.67\times {{10}^{-34}}Js$
c is the velocity of light whose value is $3\times {{10}^{8}}m{{s}^{-1}}$
$\lambda $ is the wavelength of the photon emitted.
So, we first take the minimum value of wavelength in the range of the visible spectrum, which is 400nm. The corresponding energy band gap will be the highest value in the visible spectrum range since the wavelength of light is inversely proportional to the energy. So for the minimum wavelength, we can find the bandgap as,
${{E}_{b}}=\dfrac{hc}{\lambda }=\dfrac{\left( 6.67\times {{10}^{-34}}Js \right)\times \left( 3\times {{10}^{8}}m{{s}^{-1}} \right)}{400\times {{10}^{-9}}m}$
${{E}_{b}}=3.12eV$
Next, we take the maximum value of wavelength in the range of the visible spectrum, which is 700nm. The corresponding energy bandgap will have the lowest value in the visible spectrum range since the wavelength of light is inversely proportional to the energy. So for the maximum wavelength, we can find the energy band gap as,
${{E}_{b}}=\dfrac{hc}{\lambda }=\dfrac{\left( 6.67\times {{10}^{-34}}Js \right)\times \left( 3\times {{10}^{8}}m{{s}^{-1}} \right)}{700\times {{10}^{-9}}m}$
${{E}_{b}}=1.7eV$
So, the range of the energy band gap of an LED in order to emit light in the visible region of the electromagnetic spectrum is from \[\text{1}\text{.7eV to 3}\text{.0eV}\].
So, the answer to the question is option (D).

Note: LED’s are made up of a p-n junction that needs to be in the forward-biased for the LED to work. We know that when a p-n junction is forward biased, the depletion layer which prevents the motion majority carriers from p-side to n-side and vice versa decreases due to the forward biasing, and as a result, the majority carriers from both sides can cross the junction. The majority which crosses the junction recombines with the hole or electrons, and they emit energy in the form of photons. The energy of the photon is determined by the bandgap of the LED.
Some of the materials which can be used to make LEDs are, which part of the electromagnetic they correspond to is also given:
i). Aluminium Gallium Arsenide(AlGaAs) – the infrared region of the spectrum.
ii). Gallium Arsenic Phosphide(GaAsP) – red, orange, yellow region of the visible spectrum.
iii). Aluminium Gallium Phosphide(AlGaP) – the green region of the visible spectrum.
iv). Indium gallium nitride (InGaN) – blue, blue-green, near UV.
V). Zinc Selenide(ZnSe) – blue.