Chapter 15.10, Problem 41AAP

(a)

To determine

Will the visible light be transmitted or absorbed by $\text{CdSb}$ material.

Answer to Problem 41AAP

The visible light will be transmitted by $\text{CdSb}$ material.

Explanation of Solution

Write the equation to calculate the wavelength of the radiation $\left(\lambda \right)$.

 $\lambda =\frac{hc}{E_g}$                                                                                                                   (I)

Here, energy band gap of the material is $E_g$, Planck's constant is $h$ and speed of light is c.

Conclusion:

When the critical wavelength of a material is greater than the light wavelength of $500\text{nm}$, then the material will absorb the light wave. Similarly, when the critical wavelength of the material is smaller than the light wavelength of $500\text{nm}$, then the material will transmit the light wave.

The values of Planck constant and speed of light are $6.63\times {10}^{-34}\text{J}\cdot \text{s}$ and $3\times {10}^8\text{m}/\text{s}$ respectively.

Refer Table 14.6, "Electrical properties of intrinsic semiconducting compounds at room temperature", select the energy band gap of the $\text{CdSb}$ material.

  $E_g=2.59\text{eV}$

Substitute $2.59\text{eV}$ for $E_g$, $6.63\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$ and $3\times {10}^8\text{m}/\text{s}$ for c in Equation (I).

 $\begin{array}{c}\lambda =\frac{\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{2.59\text{eV}}\\ =\frac{\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(2.59\text{eV}\times \frac{1.602\times {10}^{-19}\text{J}}{1\text{eV}}\right)}\\ =4.792\times {10}^{-7}\text{m}\\ =479.2\text{nm}\le 500\text{nm}\end{array}$

Thus, the visible light will be transmitted by $\text{CdSb}$ material.

(b)

To determine

Will the visible light be transmitted or absorbed by $\text{ZnSe}$ material.

Answer to Problem 41AAP

The visible light will be transmitted by $\text{ZnSe}$ material.

Explanation of Solution

Conclusion:

Refer Table 14.6, "Electrical properties of intrinsic semiconducting compounds at room temperature", select the energy band gap of the $\text{ZnSe}$ material.

  $E_g=2.67\text{eV}$

Substitute $2.67\text{eV}$ for $E_g$, $6.63\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$ and $3\times {10}^8\text{m}/\text{s}$ for c in Equation (I).

 $\begin{array}{c}\lambda =\frac{\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{2.67\text{eV}}\\ =\frac{\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(2.67\text{eV}\times \frac{1.602\times {10}^{-19}\text{J}}{1\text{eV}}\right)}\\ =4.649\times {10}^{-6}\text{m}\\ =464.9\text{nm}\le 500\text{nm}\end{array}$

Thus, the visible light will be transmitted by $\text{ZnSe}$ material.

(c)

To determine

Will the visible light be transmitted or absorbed by diamond material.

Answer to Problem 41AAP

The visible light will be transmitted by diamond material.

Explanation of Solution

Conclusion:

Substitute $5.40\text{eV}$ for $E_g$, $6.63\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$ and $3\times {10}^8\text{m}/\text{s}$ for c in Equation (I).

 $\begin{array}{c}\lambda =\frac{\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{5.40\text{eV}}\\ =\frac{\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(5.40\text{eV}\times \frac{1.602\times {10}^{-19}\text{J}}{1\text{eV}}\right)}\\ =2.299\times {10}^{-7}\text{m}\\ =229.9\text{nm}\le 500\text{nm}\end{array}$

Thus, the visible light will be transmitted by diamond material.