Question

A potassium chloride crystal has an energy band gap of 7.6eV above the topmost occupied band, which is full. Is this crystal opaque or transparent to light of wavelength 140 nm?

Short answer

The crystal is opaque to light of wavelength 140 nm .

Step-by-step solution

The given data

1. Band energy gap,$E_g=7.6\mathrm{eV}$
2. Wavelength of the light,role="math" localid="1661586120025" $\lambda =140\mathrm{nm}$

Understanding the concept of band gap energy

The valence band is completely occupied whereas the conduction band is not occupied. For an electron to absorb the photon energy, it should be larger than band gap. Photons with energies less than the gap width are not absorbed and the semiconductor is transparent to this radiation, whereas the photons with energies greater than the width gap are absorbed and the semiconductor is absorbed. Thus, using the given wavelength in Planck's relation, we can get the energy of the photon.

Formula:

The energy using the Planck’s relation, $E=\frac{hc}{\lambda },\mathrm{where}hc=1240\mathrm{eV}.\mathrm{nm}$ (i)

Calculation to know the crystal nature

Now, using the given wavelength of light, we get the energy of the photon as follows:

$$\begin{gathered} E_{photon}=\frac{1240\mathrm{eV}.\mathrm{nm}}{140\mathrm{nm}} \\ =8.86\mathrm{eV} \end{gathered}$$

Thus, the photon energy is greater than the given band gap energy,$8.86\mathrm{eV}>7.6\mathrm{eV}$

This implies that the light will be absorbed by theKCI crystal.

Hence, according to the concept the crystal is opaque to the light.