Question

b) A semiconductor diode laser has a peak emission wavelength of 1.55 micro meters.
Find its band gap in eV

Answer 1

Answer:

The bang gap energy of semiconductor diode laser having peak emission wavelength of 1.55 micrometers is  0.8 eV.

Explanation:

The band gap energy of semiconductor ($E_{g}$)  will be equivalent to the energy of emitted photon ($h v$ ) from the laser diode. Where 'h' is the plank's constant ($6.63 \times 10^{-34}$) and $v$ is the frequency of the photon.            

       $E_{g}=b \nu=\frac{h c}{\lambda}$      ,

       where  $\nu=\frac{ c}{\lambda}$  

                     c -   velocity of light = $3 \times 10^{8} \mathrm{~m} / \mathrm{s}$

                     λ -  wavelength of emitted photon

given,

         $\lambda=1.55 \mu \mathrm{m}=1.55 \times 16^{-6} \mathrm{~m}$

          $E_{g}=\frac{hc}{\lambda}=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{1.55 \times 10^{-6}} \mathrm{~J}$

         (  $1 eV = 1.602 \times 10^{-19} \text { joule}$ )

         $\begin{aligned}E_{g} &&=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{1.55 \times 10^{-6} \times 1.6 \times 10^{-19}} \mathrm{eV}=0.8 \mathrm{eV}\end{aligned}$

Answer 2

Given: the peak emission wavelength of the diode,λ = 1.55μm

To Find: the band-gap, E.

Solution:

To calculate E, the formula used:

• E = h x c / λ

        here, E is the band-gap energy

        h is the Planck's constant i.e. h = 6.6 x 10⁻³⁴J/Hz

        c is the velocity, i.e. 3 x 10 ⁸m/s

        λ is the emission wavelength

Applying the above formula:

E = hxc / λ

  = 6.6 x 10⁻³⁴ x 3 x 10 ⁸ / 1.55

First, convert 1.55μm into meters:

1 μm = 10⁻⁶m

1.55 μm = 1.55 x 10⁻⁶m

Putting the value in the above formula:

E  = 6.6 x 10⁻³⁴ x 3 x 10 ⁸ / (1.55 x10⁻⁶)

   = 6.6 x 3 x  10⁻³⁴x10 ⁸ x 10⁶ / 1.55

   = 19.8 x 10⁻³⁴ x 10¹⁴ / 1.55

   = 19.8 x 10⁻²⁰ / 1.55

   = 12.7 x  10⁻²⁰ J

Now convert Joules into eV:

1 J = 6.2 x 10¹⁸ eV

12.7 x  10⁻²⁰ J = 6.2 x 10¹⁸ x 12.7 x  10⁻²⁰

                      = 6.2 x 12.7 x 10⁻²

                      = 78.7 x  10⁻²

                   E = 78.7 x  10⁻²eV

Hence, the bandgap of semiconductor diode is 78.7 x  10⁻²eV.