Question

6. zinc telluride has a band gap of 2.26 ev. over what range of wavelengths of visible light is it transparent?

Answer 1

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 = $3 \times 10^8\;meters$

• Planck constant = $6.626 \times 10^{-34}\;J.s$

• Charge of proton = $1.602 \times 10^{-19}\;C$

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

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

$E = hf = \frac{hc}{\lambda}$

Where:

• E is the energy.

• h is Plank's constant.

• $\lambda$ is the wavelength.

• c is the speed of light.

Making $\lambda$ the subject of formula, we have:

$\lambda = \frac{hc}{E}$

Substituting the given parameters into the formula, we have;

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

Note: $1 \;nanometer = 1 \times 10^{-9} \;meter$

Wavelength = 550 nanometer

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