Question

For a heterojunction semiconductor laser,the bandgap of the semiconductor used is 1.44ev.By

doping,the bandgap of the semiconductor is increased by 0.2ev.Calculate the change in

wavelength of the laser


​

Answer 1

Sol: Energy gap of semiconductor, Eg = energy of emitted photon, hν

 where c = velocity of light = 3 × 108 m/s

Wavelength, λ = 1.55 μm = 1.55 × 16-6 m

Energy gap, Eg = ?



2.    Calculate the wavelength of emitted radiation from GaAs which has a band gap of 1.44 eV.

 

(Set-3–May 2008)

Sol: Energy gap of semiconductor, Eg = hν

h = Planck’s constant = 6.63 × 10-34 J-S

Answer 2

Answer:

Wavelength, λ = 1.55 μm = 1.55 × 16-6 m

Explanation:

The solution is: Energy gap of semiconductor, Eg = energy of emitted photon, hν, where c = velocity of light = 3 108 m/s

Wavelength is 1.55 m, or 1.55 16 m.

2. Determine the wavelength of the light that GaAs, with a band gap of 1.44 eV, emits.

Energy gap of a semiconductor, Eg = hν = 6.63 10-34 J-S Planck's constant

The distance over which a periodic wave's shape repeats is known as the wavelength in physics. It is a property of both travelling waves and standing waves as well as other spatial wave patterns. It is the distance between two successive corresponding locations of the same phase on the wave, such as two nearby crests, troughs, or zero crossings. The spatial frequency is the reciprocal of wavelength. The Greek letter lambda () is frequently used to represent wavelength. The term wavelength is also occasionally used to refer to modulated waves, their sinusoidal envelopes, or waves created by the interference of several sinusoids.

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