Chapter 15.10, Problem 40AAP

(a)

To determine

The minimum wavelength of the radiation absorbed by $\text{GaP}$ material.

Answer to Problem 40AAP

The minimum wavelength of the radiation absorbed by $\text{GaP}$ material is $0.552\text{μm}$.

Explanation of Solution

Write the equation to calculate the minimum 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:

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 $\text{GaP}$ material.

  $E_g=2.25\text{eV}$

Substitute $2.25\text{eV}$ 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).

 $\begin{array}{c}\lambda =\frac{\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{2.25\text{eV}}\\ =\frac{\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(2.25\text{eV}\times \frac{1.602\times {10}^{-19}\text{J}}{1\text{eV}}\right)}\\ =5.52\times {10}^{-7}\text{m}\\ =0.552\text{μm}\end{array}$

Thus, the minimum wavelength of the radiation absorbed by $\text{GaP}$ material is $0.552\text{μm}$.

(b)

To determine

The minimum wavelength of the radiation absorbed by $\text{GaSb}$ material.

Answer to Problem 40AAP

The minimum wavelength of the radiation absorbed by $\text{GaSb}$ material is $1.83\text{μm}$.

Explanation of Solution

Conclusion:

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

  $E_g=0.68\text{eV}$

Substitute $0.68\text{eV}$ 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).

 $\begin{array}{c}\lambda =\frac{\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{0.68\text{eV}}\\ =\frac{\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(0.68\text{eV}\times \frac{1.602\times {10}^{-19}\text{J}}{1\text{eV}}\right)}\\ =1.83\times {10}^{-6}\text{m}\\ =1.83\text{μm}\end{array}$

Thus, the minimum wavelength of the radiation absorbed by $\text{GaSb}$ material is $1.83\text{μm}$.

(c)

To determine

The minimum wavelength of the radiation absorbed by $\text{InP}$ material.

Answer to Problem 40AAP

The minimum wavelength of the radiation absorbed by $\text{InP}$ material is $0.98\text{μm}$.

Explanation of Solution

Conclusion:

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

  $E_g=1.27\text{eV}$

Substitute $1.27\text{eV}$ 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).

 $\begin{array}{c}\lambda =\frac{\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{1.27\text{eV}}\\ =\frac{\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(1.27\text{eV}\times \frac{1.602\times {10}^{-19}\text{J}}{1\text{eV}}\right)}\\ =9.77\times {10}^{-7}\text{m}\\ =0.98\text{μm}\end{array}$

Thus, the minimum wavelength of the radiation absorbed by $\text{InP}$ material is $0.98\text{μm}$.