Physics · Semiconductor Electronics : Materials, Devices And Simple Circuits · NEET
For an ideal LED we take the photon energy equal to the band gap: E(photon) = Eg. When a conduction-band electron drops into a valence-band hole, it loses energy equal to the gap, and that exact energy leaves as one photon. So in NEET numericals always use E = Eg. (In real devices the light can be slightly less than Eg, but that detail is not asked in NEET.)
If Eg is in electron-volts (eV) and you want lambda in nanometres (nm), use lambda(nm) = 1240 / Eg(eV). The number 1240 is just hc written in the mixed units eV.nm. Example: Eg = 2.0 eV gives lambda = 1240/2.0 = 620 nm. It saves you from plugging in h and c every time.
Same formula, different length unit. lambda in nanometres uses 1240 (lambda = 1240/Eg). lambda in angstrom uses 12400 (lambda = 12400/Eg), because 1 nm = 10 angstrom. So Eg = 1.9 eV gives 1240/1.9 = 653 nm, or 12400/1.9 = 6526 angstrom - the same length.
Method 1 (shortcut): lambda = 1240/1.9 = 653 nm. Method 2 (full SI): first turn eV to joule: E = 1.9 x 1.6e-19 = 3.04e-19 J. Then lambda = hc/E = (6.63e-34 x 3e8) / 3.04e-19 = 6.54e-7 m = 654 nm. Both give about 654 nm (red light).
A larger band gap makes blue light (shorter wavelength). Because lambda = hc/Eg, wavelength is inversely proportional to Eg. Big Eg means small lambda (blue/violet, ~450 nm); small Eg means large lambda (red, ~700 nm). This is why blue LEDs need wide-gap materials.
If you keep energy in eV, do not use hc = 6.63e-34 x 3e8 joule-metre directly, because that answer comes out in metres only after you also convert eV to joule. The clean trick is hc = 1240 eV.nm, so lambda(nm) = 1240 / Eg(eV). Mixing eV with joule-based hc without converting is the most common mistake.
An LED is constructed from a p-n junction diode using GaAsP. The energy gap is 1.9 eV. The wavelength of the light emitted will be equal to:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Eg = hc/lambda, or rearranged lambda = hc/Eg. The photon energy given out by an LED equals the band gap Eg of its material, where h is Planck's constant, c is the speed of light, and lambda is the wavelength of the emitted light.
lambda(nm) = 1240 / Eg(eV) when energy is in electron-volts and wavelength in nanometres. Use lambda(angstrom) = 12400 / Eg(eV) if the options are in angstrom.
LEDs use direct band-gap materials (like GaAs, GaAsP) where the electron energy is released mostly as light. Silicon and germanium are indirect band-gap materials that release the energy mostly as heat, so they do not glow usefully.
An LED works in forward bias. Forward bias pushes electrons and holes to the junction where they recombine, and each recombination across the gap Eg releases one photon of energy hc/lambda.
Visible light needs Eg roughly between 1.8 eV (red, ~700 nm) and 3.0 eV (violet, ~410 nm). Below about 1.8 eV the light is infrared and invisible.
