Respuesta :

Lanuel

The optimum wavelength of light for which ZnTe is transparent​ is equal to 550 nanometer.

Given the following data:

  • Energy band gap of ZnTe = 2.26 eV.

Scientific data:

  • Speed of light = [tex]3 \times 10^8\;meters[/tex]
  • Planck constant = [tex]6.626 \times 10^{-34}\;J.s[/tex]
  • Charge of proton = [tex]1.602 \times 10^{-19}\;C[/tex]

To calculate the optimum wavelength of light for which ZnTe is transparent​:

Mathematically, Einstein's equation for photon energy is given by this formula:

[tex]E = hf = \frac{hc}{\lambda}[/tex]

Where:

  • E is the energy.
  • h is Plank's constant.
  • [tex]\lambda[/tex] is the wavelength.
  • c is the speed of light.

Making [tex]\lambda[/tex] the subject of formula, we have:

[tex]\lambda = \frac{hc}{E}[/tex]

Substituting the given parameters into the formula, we have;

[tex]\lambda = \frac{6.626 \times 10^{-34}\times 3 \times 10^8}{2.26 \times 1.602 \times 10^{-19}}\\\\\lambda =\frac{1.99 \times 10^{-25}}{3.6205 \times 10^{-19}} \\\\\lambda = 5.50 \times 10^{-7}\;m[/tex]

Note: [tex]1 \;nanometer = 1 \times 10^{-9} \;meter[/tex]

Wavelength = 550 nanometer

Read more on wavelength here: https://brainly.com/question/9655595