Chapter 18, Problem 18.34QA

Interpretation Introduction

To find:

The wavelength of light emitted by the semiconducting phosphides from the band gaps values.

Answer to Problem 18.34QA

Solution:

a) ${\lambda }_{AlP} =496.4 nm$

b) ${\lambda }_{GaP} =554.0 nm$

c) ${\lambda }_{InP} =977.3 nm$

 

Explanation of Solution

1) Concept:

To find the wavelength corresponding to band gaps for semiconducting phosphides, we need to use the $E_g$ absorbed by the semiconductor.  We need to convert the given $E_g$ into Joule by using an appropriate unit conversion factor. We can calculate the wavelengthby using $E_g$ band gap and the formula.

2) Formula:

$E_g=h\times \frac{ c}{\lambda }$

3) Given:

i) $E_{g\left(AlP\right)}=241.1 \frac{kJ}{mol}$

ii) $E_{g\left(GaP\right)}=216.0 \frac{kJ}{mol}$

iii) $E_{g\left(InP\right)}=122.5 \frac{kJ}{mol}$

iv) $h=6.626 \times {10}^{-34} J.s$

v) $c= 3 \times {10}^8 \frac{m}{s}$

vi) Avogadro’s number $1 mol=6.022 \times {10}^{23} photons$

      

4) Calculations:

a) Wavelength of $AlP$:

The energy gap is the minimum energy needed to excite the electrons from the ground state for the conduction. From this energy, we can calculate the light energy needed to excite the electrons from their ground state to conduction band.

First we will calculate the energy, J/photons, by using Avogadro’s number.

$241.1 \frac{kJ}{mol}\times \frac{1000 J}{1 kJ} \times \frac{1 mol }{6.022 \times {10}^{23} photons}=4.004 \times {10}^{-19} \frac{J}{photons}$

$E_g per photon= 4.004 \times {10}^{-19} J$

Wavelength of $AlN$ is calculated as follows:

$E_g=h\times \frac{ c}{\lambda }$

$\lambda = \frac{h\times c}{E_g}$

$\lambda = \frac{\left( 6.626 \times {10}^{-34} J.s\right) \times \left(3 \times {10}^8 \frac{m}{s}\right)}{4.004 \times {10}^{-19} J} = 4.964 \times {10}^{-7} m$

$\lambda =4.964 \times {10}^{-7} m \times \frac{1 \times {10}^{9}nm}{1 m} =496.4 nm$

$\lambda =496.4 nm$

b) Wavelength of $GaP$:

First we will calculate the energy, J/photons, by using Avogadro’s number.

$216.0 \frac{kJ}{mol}\times \frac{1000 J}{1 kJ} \times \frac{1 mol }{6.022 \times {10}^{23} photons}=3.59 \times {10}^{-19} \frac{J}{photons}$

$E_g per photon= 3.59 \times {10}^{-19} J$

Wavelength of $AlN$ is calculated as follows:

$E_g=h\times \frac{ c}{\lambda }$

$\lambda = \frac{h\times c}{E_g}$

$\lambda = \frac{\left( 6.626 \times {10}^{-34} J.s\right) \times \left(3 \times {10}^8 \frac{m}{s}\right)}{3.59 \times {10}^{-19} J} = 5.54 \times {10}^{-7} m$

$\lambda =5.540 \times {10}^{-7} m \times \frac{1 \times {10}^{9}nm}{1 m} =554.0 nm$

$\lambda =554.0 nm$

c) Wavelength of $InP$:

First we will calculate the energy, J/photons, by using Avogadro’s number.

$122.5 \frac{kJ}{mol}\times \frac{1000 J}{1 kJ} \times \frac{1 mol }{6.022 \times {10}^{23} photons}=2.034 \times {10}^{-19} \frac{J}{photons}$

$E_g per photon= 2.034 \times {10}^{-19} J$

Wavelength of $AlN$ is calculated as follows:

$E_g=h\times \frac{ c}{\lambda }$

$\lambda = \frac{h\times c}{E_g}$

$\lambda = \frac{\left( 6.626 \times {10}^{-34} J.s\right) \times \left(3 \times {10}^8 \frac{m}{s}\right)}{2.034 \times {10}^{-19} J} = 9.773 \times {10}^{-7} m$

$\lambda =9.773 \times {10}^{-7} m \times \frac{1 \times {10}^{9}nm}{1 m} =977.3 nm$

$\lambda =977.3 nm$

Conclusion:

The wavelength is calculated from the energy gaps present in the semiconductors and Avogadro’s value.